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MATHEMATICS
QUARTER 3
DO_Q3_MATHEMATICS_10_MODULE1-9
MATHEMATICS – Grade 10
Alternative Delivery Mode
Quarter 3 – Modules
Second Edition, 2021
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Published by the Department of Education
Development Team of the Module
Writers: Oliver G. Mariano, Augusto B. Logronio, Irene C. Imperial,
Fidencio C. Carreon Jr., Leny M. Hilario, Eliza C. Sorianosos,
Midjet V. Barbosa, Niel Raymond K. Factor, Resty B. Satuito
Content Editors: Augusto B. Logronio, Emma D. Tejuco, Glaiza M. Batas,
Irene C. Imperial, Leny M. Hilario, Midjet V. Barbosa, Resty B. Satuito
Consultant: Dr. Rudy Fran Falcunitin
Language Editors: John Paul E. Licoan
Illustrator: Shella Marie A. Reyes
Layout Artist: Resty B. Satuito, Raphael A. Lopez
Management Team:
MELITON P. ZURBANO, Assistant Schools Division Superintendent
(OIC-SDS)
FILMORE A. CABALLERO, CID Chief
JEAN A. TROPEL, Division EPS In-Charge of LRMS
MARILYN B. SORIANO, Division Mathematics Coordinator
Printed in the Philippines by ________________________
Department of Education – National Capital Region – SDO VALENZUELA
Office Address:
Telefax:
E-mail Address:
Pio Valenzuela St., Marulas, Valenzuela City
(02) 292 – 3247
sdovalenzuela@deped.gov.ph
What I Need to Know
This module was designed and written to help you, students, to identify
situations that illustrate permutations. The scope of this module permits you to learn
over any condition, especially in this pandemic situation.
After going through this module, you are expected to illustrate the
permutation of objects (M10SP-IIIa-1).
What I Know
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. Which of the following refers to an arrangement in which order is important?
A. combination
B. permutation
C. statistics D. combinatorics
2. Which of the following refers to the product of all the positive integers from 1
to 𝑛?
A. 𝑛-series
B. 𝑛-factors
C. 𝑛 factorial
D. 𝑛 multiples
3. What is the value of 𝟎!?
A. 1
B. 0
C. -1
D. undefined
𝑛!
4. If 𝑛 = 6, what is the value of 2!?
A. 1440
B. 720
C. 360
D. 180
5. Given that P (8, r) = 336, what is the value of 𝑟?
A. 6
B. 5
C. 4
D. 3
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of the lesson and proceed to the activities or exercises.
Lesson
1
Illustrating Permutations
In this lesson, you will learn about how these arrangements can be possibly
made as well as the number of ways in which they can be done. Further, you will be
able to identify particular situations that exhibit such arrangements, which we call
as Permutations.
What’s New
Angie, Belle, Carl, Darren, Edgar, and Francine are to stand in a row for
picture taking. How many arrangements are possible such that one may stand
next to any other. How many choices of persons do you have for the first position
in the row? second? Last? What’s concept is needed to solve the problem?
What is it
Imagine yourself doing the following activities, which of them do you think
requires you to do the activity in some specific order?
1. In an ice cream stand, you need to select three of the five available ice cream
flavors to eat which are strawberry, chocolate, mango, ube, and cheese.
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DO_Q3_Mathematics10_Lesson1
2. You bought one pack of each of the following fruits: strawberry, pomelo, and
grapes. You want to store two of them into canisters of colors green and
orange.
3. Your teacher asked you to name all the subsets of the set {𝑥, 𝑦, 𝑧}.
4. You are the leader of a group, and you need to pick an assistant leader and a
secretary for your group.
In the activities presented above, some of them simply ask you to choose while
others require you not only to pick but also to arrange them in some ways. In these
activities, the arrangements that we do wherein the order in which the objects are
arranged is called Permutation.
The situation presented in numbers 2 and 4 illustrate permutation. In number 4,
for instance, assigning person A as your assistant and person B as the secretary is
completely different when you assign person B as your assistant and person A as the
secretary. Clearly, in permutation of objects, order is important.
However, the situations given in numbers 1 and 3 do not illustrate
permutations. In the case of the activity given in number 3, regardless of the order
in which the elements of the given set are selected to form a subset, it gives us the
same subset. Particularly, you may recall that the set {𝑥, 𝑦} is the same as the set
{𝑦, 𝑥}.
Illustrating Permutations
Let us further deepen our understanding of permutations through the
following illustrations.
Illustration 1: Consider the situation presented in number 2 earlier which
is “You bought one pack of each of the following fruits: strawberry, pomelo, and
grapes. You want to store two of them into canisters of colors green and orange.”
Let us try to exhaust all the possible arrangements that can be done in this
situation.
Because there are three different fruits and only two of them can be stored in
canisters, there is always one fruit which will be left out in any of our arrangements.
To begin, first, let’s say that we will not store the grapes in a canister. The table below
shows how we can place the strawberries and pomelo in the canisters.
Green Canister
Orange Canister
Strawberry
Pomelo
Pomelo
Strawberry
Thus, we can see that there are only two ways in which we can store the
strawberries and pomelo in the canisters. Now, suppose we leave out the pomelo.
Then, we will have the following arrangements for strawberries and grapes.
Green Canister
Orange Canister
Strawberry
Grapes
Grapes
Strawberry
Clearly, there are also two ways in which we can store the strawberries and
grapes in the canisters. Lastly, let us try leaving out the strawberries. Then we can
see from the table below that we only have the following arrangements.
Green Canister
Grapes
Pomelo
Orange Canister
Pomelo
Grapes
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DO_Q3_Mathematics10_Lesson1
Hence, we only have two ways to store them. To sum it up, there are only 6
ways to store 3 fruits in a canister taken 2 at a time. This situation illustrates the
permutation of 3 objects taken 2 at a time.
Illustration 2: Let us consider another situation, suppose you want to use a
4-digit passcode for securing access to a certain application in your cellphone. You
want this passcode to contain exactly one of each of the digits 1, 2, 5, and 9. Let us
analyze all the possible passcodes that you may use.
First, we establish that this situation involves permutation. Clearly, passcodes
are generated according to a certain order, a different arrangement of the digits will
not guarantee access to the application in your cellphone. By listing, we can generate
the following passcodes that you may use.
1259
2159
5129
9125
1295
2195
5192
9152
1529
2519
5219
9215
1592
2591
5291
9251
1925
2915
5912
9512
1952
2951
5921
9521
Notice that even if all these arrangements contain the same digits, they are
known to be distinct from one another. Hence, there are 24 ways in which we can
arrange 4 objects taken 4 at a time.
Generally, if you have 𝑛 objects and you want to arrange them 𝑟 at a time,
then you have the permutation of 𝑛 objects taken 𝑟 at a time, where 𝑟 ≤ 𝑛.
Factorial Notation
The product of a positive integer 𝑛 and all the positive integers less than it is
called 𝒏 factorial. In symbols, we write 𝒏! = 𝑛(𝑛 − 1)(𝑛 − 2) ⋅ … ⋅ 3 ⋅ 2 ⋅ 1. Further, we
define 0! = 1.
With this definition, we can say that 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 and 6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅
1 = 720. To illustrate this further, let us have the following:
Illustration 3:
Determine the value of the following expressions containing factorials.
a.
b.
7!
7⋅6⋅5!
= 5! = 7 ⋅ 6 = 42
(7−2)!
9!
9⋅8⋅7⋅6⋅5⋅4!
=
=9⋅8⋅7⋅5
3!4!
6⋅4!
= 2 520
The Permutation of 𝒏 objects taken 𝒓 at a time
Consider the situation in number 4, that is, “You are the leader of a group,
and you need to pick an assistant leader and a secretary for your group.” Assume
that you have 𝑛 members and we are to take 𝑟 of them at a time, where 𝑟 = 2 (since
only two positions are to be assigned). Then, we have,
𝑃(𝑛, 𝑟) = 𝑛(𝑛 − 1) = 𝑛(𝑛 − 2 + 1)
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DO_Q3_Mathematics10_Lesson1
Extending this concept for other values of 𝒓, we have,
𝐏(𝐧, 𝐫)
Equivalent Expression
𝑛(𝑛 − 1)(𝑛 − 2)
𝑛(𝑛 − 1)(𝑛 − 3 + 1)
𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)
𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 4 + 1)
𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4)
𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 5 + 1)
Using these specific instances, let us try to derive the formula for finding the
permutation of 𝑛 objects taken 𝑟 at a time. As observed, we can say in general that,
𝑃(𝑛, 𝑟) = 𝑛(𝑛 − 1)(𝑛 − 2) ⋅ … ⋅ (𝑛 − 𝑟 + 1).
Without loss of generality, this is the same as
𝑛(𝑛 − 1)(𝑛 − 2) ⋅ … ⋅ (𝑛 − 𝑟 + 1)(𝑛 − 𝑟)(𝑛 − 𝑟 − 1) ⋅ … ⋅ 3 ⋅ 2 ⋅ 1
𝑃(𝑛, 𝑟) =
(𝑛 − 𝑟)(𝑛 − 𝑟 − 1) ⋅ … ⋅ 3 ⋅ 2 ⋅ 1
Using the definition of factorial, we can say that,
𝑛!
𝑃(𝑛, 𝑟) =
𝑟≤𝑛
(𝑛 − 𝑟)!
The last equation gives the permutation of 𝑛 objects taken 𝑟 at a time.
𝐫
3
4
5
Observe that when 𝑛 = 𝑟, 𝑃(𝑛, 𝑟) = 𝑛𝑃𝑛 =
𝑛!
(𝑛−𝑛)!
𝑛!
= 0! = 𝑛!
Study the illustrations below to see how this formula may be applied.
Illustration 4:
Evaluate the following permutations.
a. 𝑃(4,2) =
4!
(4−2)!
4!
= 2! =
4⋅3⋅2!
2!
= 4 ⋅ 3 = 12
Illustration 5:
Solve for the unknown in each of the following.
a. 𝑃(𝑛, 3) = 60
b. 𝑃(7, 𝑟) = 210
𝑛!
= 60
(𝑛 − 3)!
7!
= 210
(7 − 𝑟)!
𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)!
= 60
(𝑛 − 3)!
7⋅6⋅5⋅4⋅3⋅2⋅1
= (7 − 𝑟)!
210
𝑛(𝑛 − 1)(𝑛 − 2) = 60
24 = (7 − 𝑟)!
𝑛(𝑛 − 1)(𝑛 − 2) = 5 ⋅ 4 ⋅ 3
4! = (7 − 𝑟)!
𝑛=5
4=7−𝑟
𝑟=3
What’s More
A. Write P if the give situation illustrates a permutation and NP if it does
not.
1.
2.
3.
4.
5.
Determining the finalists in a singing contest
Forming lines given 7 points, no three of which are collinear
Assigning a locker to each of the 8 athletes
Entering the PIN code in an ATM
Electing homeroom officers
B. Evaluate the following expressions.
11!
9!
1. 2!5!
2. (9−5)!
3. (7 − 1)! 2!
4
4. 𝑃(8,5)
5. 𝑃(10,6)
DO_Q3_Mathematics10_Lesson1
What I can Do
Do as indicated.
1. Find the value of 𝑛 in the equation 𝑃(𝑛, 4)= 3 024.
2. Determine the value of 𝑟 in the equation 𝑃(12, 𝑟) = 1 320.
3. Give the permutation statement involved in each situation. Do not solve the
problem.
a. A plantito wants to arrange all his 7 pots in a row.
b. A talent scout picks two of the 5 auditionees, one for a lead role and one
for a supporting role.
c. A basketball coach selects 5 players from his team for the different
positions in the game.
Assessment
Directions: Write the letter of the correct answer on a separate sheet of paper.
For items 1 to 3, determine the value of each expression.
1. (5 − 1)! 2!
A. 24
6!
2. (6−3)!
A. 360
3.
B. 48
C. 60
D. 120
B. 240
C. 120
D. 60
8!
2!3!
A. 3 360
B. 1 680
C. 1 120
D. 56
4. Which of the following gives the number of permutations of 𝒏 objects taken 𝒓
at a time?
𝑛!
𝑛!
𝑛!
𝑛!
A. 𝑟!
B. 𝑛−𝑟!
C. 𝑟!(𝑛−𝑟)!
D. (𝑛−𝑟)!
5. If 𝑃(𝑛, 4) = 5 040, what is the value of 𝒏?
A. 10
B. 9
C. 8
D. 7
At this point, you are already familiar in arranging objects in different ways.
This lesson on Word Problems Involving Permutations will extend the knowledge that
you have so far. It will discuss about how to solve real word problems involving
permutations.
After going through this module, you are expected to solve problems involving
permutations (M10SP-IIIb-1).
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DO_Q3_Mathematics10_Lesson2
What I Know
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. Evaluate the expression 𝑃(10,3).
A. 720
B. 504
C. 120
D. 30
2. How many permutations of the letters a, b, c, and d are there?
A. 24
B. 12
C. 6
D. 3
3. An art gallery has 8 paintings to hang and 3 vacant positions, each of which
will hold 1 painting. In how many ways can these 3 locations be filled by the
paintings?
A. 24
B. 60
C. 210
D. 336
4. Find the number of permutations of the letters in “ABNKKBSNAKO”.
A. 831,200
B. 831,600
C. 1,663,200
D. 4,435 ,200
5. In how many ways can 8 different books be arranged on a shelf?
A. 40, 320
B. 5760
C. 5040
D. 8
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
Lesson
2
Word Problems Involving
Permutations
There are situations where choices are to be made. And in some case the order
in which the choices are made is important. For example, in how many ways can a
president and a vice-president be elected from a group of 52 students?” Express your
answer in factorial notation.
The situation given above is called Permutation. Permutation is an
arrangement of objects in some specified order.
Permutation of Distinct Objects
Suppose you want to form 2-digit numbers from the digits 1,2,3, and 4 under
the condition that no digit is to be used twice. See the illustration below.
12, 13, 14
21, 23, 24
31, 32, 34
Observe that there is an order in the way the digits are placed,
since you are required to arrange the digits two at a time from
the given 4 digits. This situation can be translated as
permutation of four objects taken two at a time, in symbol 4P2.
41, 42, 43
In general, 𝑃(𝑛, 𝑟) denotes the permutation of n objects taken r at a time. There
are some notations for 𝑃(𝑛, 𝑟), these are 𝑛𝑃𝑟 or 𝑃𝑟𝑛 . The formula can be written as
𝑷(𝒏, 𝒓) =
𝒏!
,
(𝒏−𝒓)!
where 𝟎 ≤ 𝒓 ≤ 𝒏.
In the given problem above n = 4 and r = 2.
=
4(3)(2)(1)
2(1)
= 12 ways
6
There are 12 numbers formed
when the four digits are
arranged 2 digits at a time.
DO_Q3_Mathematics10_Lesson2
4!
.
(4−2)!
Hence, 𝑃(4,2) =
=
4!
2!
Example 1: In how many ways can an advertising company promote 8 items 4 at a
time during a 3- minute commercial period at TV time!
Solution: n = 8 and r = 4
𝑃(8,4) =
=
8!
(8−4)!
=
8!
4!
8(7)(6)(5)(4!)
4!
You may stop expanding the factorial
notation up to 4! to cancel out 4! in
the denominator
= 1680 ways
Example 2: In how many ways can 3 boys and 4 girls be seated in a row of 7 seats
if girls insist on sitting next to each other?
Solution: First you may consider of sitting a group of girls in a row followed by boys.
Then, a boy may occupy first from the row followed by the group of girls and so on
𝑔1 𝑔2 𝑔3 𝑔4 𝑏1 𝑏2 𝑏3
𝑏1 𝑔1 𝑔2 𝑔3 𝑔4 𝑏2 𝑏3
𝑏1 𝑏2 𝑔1 𝑔2 𝑔3 𝑔4 𝑏3
𝑏3 𝑏1 𝑏2 𝑔1 𝑔2 𝑔3 𝑔4
Observe that girls may occupy the first from the right
of the available 4 positions considering that girls sit
always next to each other, and may also occupy the
2nd, 3rd and 4th positions knowing that they are
grouped together while boys are occupying the other
3 positions.
Thus, the arrangement you will consider is 𝐺 𝑏1 𝑏2 𝑏3 , where 𝐺 is the group of girls.
Therefore, in that situation the formula is
𝑃(4,4) =
4!
(4−4)!
= 4! = 24
Note: This is just one of the considerations, remember
that girls may rearrange themselves.
The number of ways where girls may rearrange themselves but still, they are
4!
occupying the consecutive seats will be 𝑃(4,4) = (4−4)! = 𝟐𝟒 . Since there are four seats
available for them and all of them may sit anywhere as long as they are grouped
together.
Therefore, the total number of ways of arranging 4 girls and 3 boys in a row,
where girls sit next to each other is
4! 4! = (24)(24) = 576 𝑤𝑎𝑦𝑠
Permutation of Identical Objects
Now consider another instance when you are asked to arrange identical
objects. For example, you want to find different 4- digit numbers which can be formed
using the digits in 3665.
Solution:
If all the digits are distinct, you should be able to find 𝑃(4,4) = 24.
But since the digits are not distinct it is not enough to use the first formula.
You may denote the 6s as 61 and 62. Now, from the 24 numbers formed in which the
6s have exchanged places,
361625 and 362615
7
DO_Q3_Mathematics10_Lesson2
you remove the subscript. You will get 3665 as identical numbers. Apply the same
concept of removing subscript in
613625 and 623615
Notice that you have the same number 6365. Hence, we do not have 24
different numbers. You can only get half of 24 numbers which is 12.
This situation is called permutation of identical objects. To find the
distinguishable permutation of identical objects use the formula
𝑷=
𝒏!
𝒏𝟏 !𝒏𝟐 !𝒏𝟑 !…𝒏𝒌 !
, where n is total number of objects and n1, n2, n3, …, nk are
numbers of identical elements. In the problem above, the values of n = 4 and n1=2
(two 6), n2 = 1 (one 3) and n3 = 1(one 5) can be substituted in the formula.
4!
𝑃 = 2!1!1! = 12 different 4- digit numbers
Example 1: How many different permutations can be made from the word
“MASAYA”?
Solution: n = 6, n1 = 3, n2 = 1, n 3 = 1, and n4 = 1
6!
𝑃(𝑀𝐴𝑆𝐴𝑌𝐴) =
There are 6 letters, number of A ‘s = 3 and the
3! 1! 1! 1!
= 𝟏𝟐𝟎 ways
number of letters M, S and Y = 1.
Example 2: Robert plans to visit his grandma’s house which is located at the other
street in their village. He always walks south or east direction. Below is an illustration
of the street in their village. In how many routes from Robert’s house to Grandma’s
house?
Robert’s House
7!
𝑃(𝑆𝑆𝐸𝐸𝐸𝐸𝐸) =
2! 5!
= 21 routes
Grandma’s House
To go from Robert’s house to Grandma’s house, he must walk 2 blocks
south (𝑺) and 5 blocks east (𝑬).
Permutation of Circular Objects
Amy. Bella and Carla would like to sit in a round table during the birthday
party of their friend Dindo. Let us look at the possible ways where these girls may
arrange themselves in a circular table. In how many possible ways can they be
seated? Which of the following arrangement are distinct and are the same?
Amy
Carla
Table
1
Bella
Bella
Amy
Table
4
Bella
Table
2
Carla
Table
3
Carla
Amy
Bella
Carla
Bella
Carla
Amy
Table
5
Table
6
Amy
Bella
Amy
Carla
Look at who is on the right and on the left of each girl. Notice that when
you rotate the names in clockwise direction, there are arrangements
which coincides with the other.
8
DO_Q3_Mathematics10_Lesson2
This kind of arrangement is called permutation of circular objects. In the
given problem observe that Table 1, Table 2, and Table 3 represent the same
arrangement. Whereas, in Table 4, Table 5, and Table 6 show another set of
arrangement. The distinct circular permutation is (3 – 1)! = 2! = 2(1) = 2.
In general, for 𝒏 distinct objects, there are (𝒏 − 𝟏)! ways of arranging them in
a circle. The arrangement can be done by fixing one object in one place and the other
objects are permuted as shown below.
Amy
Table 1
Carla
Amy
Table 4
Bella
Carla
Bella
Example 1: In how many ways can 5 children join hands to form a ring?
Solution: 𝑛 = 5
so, (𝑛 – 1)! = (5– 1)! = 4! = 24 𝑤𝑎𝑦𝑠
Example 2: In how many ways can John, Michael, Lea and Sarah be seated at a
circular table if Lea and Sarah insist on sitting next to each other?
Solution:
Here we can fix the girls in one place and two boys may rearrange themselves.
There are (3 − 1)! = 𝟐! ways in these arrangements. But the girls may also shift
their positions, thus there are 𝟐! ways.
By multiplication rule there are 𝟐! 𝟐! = 4 ways of arrangements.
A. Solve the following problems.
1. In how many ways can 4 runners win first, second and third places?
2. In how many ways can a chairman, vice chairman and secretary be elected
from 15 members if no person may hold more than one position?
3. In how many ways can 4 boys and 5 girls be seated in a row of 9 seats:
a. if they may sit anywhere?
b. if boys insist on sitting next to each other?
B. Solve.
1. Eight people apply for 3 job available positions. In how many different ways
can the 3 positions be filled if the positions are all different?
2. Due to a pandemic limited people are allowed to work in a certain cafeteria.
The 12 workers in cafeteria crew rotate among 3 kinds of jobs. In how many
ways can the crew be assigned the jobs of 2 cooks, 7 servers, and 3
Solve the following problems.
1. In how many ways can a club with 20 members choose a president, a vice
president, a secretary, a treasurer, and an auditor?
2. In how many ways can 4 people be seated in a row of 12 chairs?
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DO_Q3_Mathematics10_Lesson2
3. A standard deck of plating cards consists of 53 cards, with 13 cards in each
four suits (clubs, spades, diamonds, and hearts). How many ways are there
to deal 13 cards if the order in which the cards are dealt is important?
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. A2D company advertises two job openings, one for news reporter and one for
a camera man. If 12 people are qualified for either position apply, in how
many ways can the opening be filled?
A.10
B. 12
C. 66
D. 132
2. If 𝑃(𝑛, 2) = 42, find 𝒏.
A.6
B. 7
C. 8
D. 9
3. How many different 4-digit numbers can be formed using the digits in 3573?
A. 24
B. 18
C. 12
D. 10
4. In how many ways can 8 people be seated in row of 8 seats if two them insist
on sitting next to each other?
A. 40, 320
B. 20, 160
C. 10, 080
D.84
5. In how many ways can 6 people sit at a round table?
A. 720
B. 120
C. 56
D. 30
This module was designed and written to help students illustrate the
combination of objects. The scope of this module permits the students to learn over
any condition, especially in this pandemic situation.
After going through this module, you are expected to illustrate the
combination of objects. (M10SP-IIIc-1)
What I Know
Directions: Write the letter of the correct answer on a separate sheet of paper.
For items 1 – 2. A committee of five people is to be chosen from a group of six men
and four women.
1. How many committees are possible if there are three men and two women?
A. 120
B. 150
C. 300
D. 3600
2. How many committees are possible if there is to be majority of women?
A. 33
B. 66
C. 109
D. 360
3. In how many ways can you pick 5 cards if you must choose a queen, then a
king and then 3 other cards if every card is drawn one at a time?
A. 1,881,600
B. 2,598,960
C. 3,292,800
D. 41,583,360
4. A box contains 8 blue socks and 6 red socks. Find the number of ways two
socks can be drawn from the box if they must be the same color.
A. 15
B. 28
C. 43
D. 420
5. How many diagonals does a 16-sided polygon have?
A. 104
B. 120
C. 224
D. 240
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
10
DO_Q3_Mathematics10_Lesson3
Lesson
3
Combination of Objects
Have you ever wondered how many combinations you need to bet to win a
Grand Lotto 6/55? In the previous lessons, we have been concerned with ordered
arrangements of elements of a set. This module will help you illustrate combination
of objects where order of selection or arrangement is not important.
From our previous lessons we have been concerned with ordered arrangements
of elements of a set. Our discussion now pertains to subset of a set without regard
to the relative order of the elements in the subsets.
If a committee of five people be formed from a group of 9 people?
1. Does the order of people matter in our given? Explain your answer
2. In how many ways can a committee of five people be formed from a group of 9
people? Show your solution.
Consider the number of ways a reader can choose three (3) books from four (4)
math books (Algebra, Geometry, Trigonometry and Statistics). Unlike positioning
books on a library shelf, order does not matter. Choosing books A, G, and T is the
same as choosing G, A, and T, and so on.
Combination of 3 books
Permutation of each combination
AGT
AGT, ATG, GAT, GTA, TAG, TGA
AGS
AGS, ASG, GSA, GAS, SAG, SGA
ATS
ATS, AST, TSA, TAS, SAT, STA
GTS
GTS, GST, TGS, TSG, SGT, STG
4 combinations
24 permutations
When order matters, there are 4P3, or 24 ways to choose 3 books from 4 books.
There are 3! or 6 ways to choose the same 3 books. So, the number of
combinations is
24
3!
24
= 3∙2∙1 = 4 ways.
Definition. A combination of 𝑛 distinct objects taken 𝑟, 0 ≤ 𝑟 ≤ 𝑛, at a time is a subset
containing 𝒓 objects of a set that has 𝒏 objects.
The notation we use for the number of combinations of 𝑛 elements taken 𝑟 at
a time is 𝑪(𝒏, 𝒓). Other symbols for this number are nCr, 𝒏𝑪𝒓, 𝑪𝒏,𝒓 , 𝑪𝒏𝒓 , and (𝒏𝒓).
Combination of Different Objects
The number of combinations of 𝑛 distinct objects taken 𝑟, 0 ≤ 𝑟 ≤ 𝑛, at a time, denoted
by 𝐶(𝑛, 𝑟), is 𝑪(𝒏, 𝒓) =
𝒏!
𝒓!(𝒏−𝒓)!
.
11
DO_Q3_Mathematics10_Lesson3
In general, 𝑟 objects can be chosen from 𝑛 distinct objects in 𝑃(𝑛, 𝑟) ways and
𝑟 objects can be arranged in 𝑟! ways. So, the number of combinations of 𝑛 distinct
objects taken 𝑟 at a time is:
𝑛!
𝑃(𝑛, 𝑟) (𝑛 − 𝑟)!
𝑛!
𝐶(𝑛, 𝑟) =
=
=
𝑟!
𝑟!
𝑟! (𝑛 − 𝑟)!
Example 1. In how many ways can a committee of 7 people be formed from a group
of 12 people?
[Solution] Since we are only concerned with a subset having 7 people (𝒓) of a set
with 12 people (𝒏), the number of possible committees is then,
Example 2. Using an ordinary deck of cards, in how many ways can you draw a
poker hand (5 cards) of the same suit?
Note that an ordinary deck has four suits, each
[Solution] 𝒏 = 𝟏𝟑, 𝒓 = 𝟓
of which has thirteen cards. Thus, if we only
13!
consider one suit, we may draw 5 cards from
𝐶(13,5) =
= 1287 𝑤𝑎𝑦𝑠
this suit.
5! 8!
But since there are four suits, then this can be done in 4(1287) = 𝟓𝟏𝟒𝟖 𝒘𝒂𝒚𝒔.
Example 3. Let 𝑆 be the collection of eight points in the plane with no three of which
are collinear. How many polygons can be possibly formed that have points from 𝑆 as
vertices?
[Solution]
The polygon may have
𝑋 = 𝐶(8,3) + 𝐶(8,4) + 𝐶(8,5) + 𝐶(8,6) + 𝐶(8,7) + 𝐶(8,8)
3,4,5,6,7, or 8 vertices. Let
𝑿 be the number of
possible polygons.
8!
8!
8!
8!
8!
8!
=
+
+
+
+
+
3! 5! 4! 4! 5! 3! 6! 2! 7! 1! 8! 0!
= 56 + 70 + 56 + 28 + 8 + 1 = 𝟐𝟏𝟗 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒑𝒐𝒍𝒚𝒈𝒐𝒏𝒔
Example 4. A committee of 7 members is to be chosen from six (6) Grade 10
students, four (4) Grade 11 students and five (5) Grade 12 students. In how many
ways can this be done if there must be at least one member from each grade level
and at least 3 Grade 10 students in the committee.
[Solution] The various methods of selecting the persons from various groups are
shown below:
Committee of 7 members
Cases
Grade 10
Grade 11
Grade 12
Case 1
3
2
2
Case 2
3
3
1
Case 3
3
1
3
Case 4
4
2
1
Case 5
4
1
2
Case 6
5
1
1
The number of ways of choosing the committee members
= 𝐶𝑎𝑠𝑒 1 + 𝐶𝑎𝑠𝑒 2 + 𝐶𝑎𝑠𝑒 3 + 𝐶𝑎𝑠𝑒 4 + 𝐶𝑎𝑠𝑒 5 + 𝐶𝑎𝑠𝑒 6
6×5×4
4×3
5×4
)( )( )
3×2×1
2×1
2×1
6×5×4
4×3×2
5
(3×2×1) (3×2×1) (1)
6×5×4
4
5×4×3
(3×2×1) (1) (3×2×1)
Case 1: 𝐶(6,3) ∙ 𝐶(4,2) ∙ 𝐶(5,2) = (
= (20)(6)(10) = 1,200
Case 2: 𝐶(6,3) ∙ 𝐶(4,3) ∙ 𝐶(5,1) =
= (20)(4)(5) = 400
Case 3: 𝐶(6,3) ∙ 𝐶(4,1) ∙ 𝐶(5,3) =
12
= (20)(4)(10) = 800
DO_Q3_Mathematics10_Lesson3
6×5
4×3
5
) ( ) ( ) = (15)(6)(5) = 450
2×1
2×1
1
6×5
4
5×4
( ) ( ) ( ) = (15)(4)(10) = 600
2×1
1
2×1
6
4
5
(1) (1) (1) = 120
Case 4: 𝐶(6,4) ∙ 𝐶(4,2) ∙ 𝐶(5,1) = (
Case 5: 𝐶(6,4) ∙ 𝐶(4,1) ∙ 𝐶(5,2) =
Case 6: 𝐶(6,5) ∙ 𝐶(4,1) ∙ 𝐶(5,1) =
Therefore, the total number of ways = 1200 + 400 + 800 + 450 + 600 + 120 = 𝟑, 𝟓𝟕𝟎 .
A. Solve the following.
1. Let 𝑀 be the collection of eight points in the plane with no three points
are collinear. Find the number of triangles that have points of 𝑀 as
vertices.
2. A person has 12 friends whom 8 are relatives. In how many ways can
he invite 7 guests such that 5 of them are relatives?
3. A box contains 7 red, 6 white and 4 blue balls. How many selections of
three balls can be made so that
(a) all three are red? (b) none is red?
(c) one is of each color?
4. If 𝑃(10, 𝑟) = 604,800 and 𝐶(10, 𝑟) = 120, find the value of 𝑟.
5. Find the number of ways of selecting 4 letters from the word
"𝐸𝑋𝐴𝑀𝐼𝑁𝐴𝑇𝐼𝑂𝑁".
B. Solve the following.
1. Ten points are on the circumference of a circle where no three points
are collinear, how many quadrilaterals can be formed?
2. A class contains 10 students with 6 men and 4 women. Find the
number of 𝑛 ways to:
a. select a 4-member committee from the students.
b. select a 4-member committee with 2 men and 2 women.
3. Find the number 𝑚 of committees of 5 with a given chairperson from
12 people.
4. A woman has 11 close friends. Find the number of ways she can invite
5 of them to dinner where:
a. there are no restrictions.
b. two of her friends are couple and they want to attend together.
Solve the following problems.
1. A box has 6 blue socks and 4 white socks. Find the number of ways two
socks can be drawn from the box where:
a. there are no restrictions.
b. they are different colors.
c. there are the same color.
2. A committee of 3 is selected from 5 girls and 7 boys. How many numbers
of committees can be formed if:
a. there are no restrictions?
b. there are all boys?
c. there are 2 boys and 1 girl?
d. there’s at least 1 girl?
e. there are at least 2 boys?
13
DO_Q3_Mathematics10_Lesson3
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. Which expression is NOT equivalent to 𝐶(7,5)?
7!
𝑃(7,5)
A. 5!(7−5)!
B. 𝐶(7,2)
C. 5!
D. (57)
2. How many ways can two slices of pizza be chosen from a plate containing one
slice each of pepperoni, sausage, mushroom, and cheese pizza?
A. 2
B. 6
C. 8
D. 12
3. Twelve points are located on the circumference of a circle. Lines are drawn to
connect all possible pairs of points. How many lines are drawn?
A. 24
B. 66
C. 132
D. 144
4. How many committees of 4 men and 7 women can be selected from a group of 9
men and 10 women?
A. 𝐶(9,4) 𝐶(10,7)
B. 𝑃(9,4) ∙ 𝑃(10,7) C. 19C11
D.𝐶(9,4)+ 𝐶(10,7)
5. Filipino restaurant serves dinner for 8 people consists of 3 items from column A,
4 items from column B and 3 items from column C. If columns A, B and C have
5, 7 and 6 items respectively how many different dinner combinations are
possible?
A. 65
B. 1020
C. 7000
D. 43,758
This module was designed and written to help you, students, to extend your
ideas on permutations and combinations for you to be able to differentiate
permutation from combination to solve related problems.
After going through this module, you are expected to differentiate permutation
from combination of 𝒏 objects taken 𝒓 at time (M10SP-IIIc-2).
What I Know
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. A coin is tossed 3 times. Find out the number of possible outcomes.
A. 1
B. 2
C. 6
D. 8
2. In how many ways can the letters of the word 'VANGIE' be arranged such that
the vowel must occupy only the odd positions?
A. 36
B. 64
C. 120
D. 720
3. There are 5 men and 7 women in a group. A committee of 3 men and 2 women is
to be formed from the group. In how many ways can the committee be formed?
A. 5
B. 6
C. 210
D. 2520
4. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9
which are divisible by 5 and none of the digits is repeated?
A. 8
B.16
C. 20
D. 24
5. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which
are divisible by 5 and digits can be repeated?
A. 20
B. 25
C. 30
D. 36
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
14
DO_Q3_Mathematics10_Lesson4
Lesson
4
Differentiate Permutation from Combination of
n objects taken r at a time
The ideas and skills that you have learned in the previous lessons regarding
permutation and combination will further help you in forming conclusions and
making decisions in identifying whether a question is a combination or permutation
problem. Moreover, it will also help you with your difficulty of knowing the difference
between combinations and permutations (M10SP-IIIc-2). To be able to this, perform
each activity that follows. Seek the assistance of your teacher and peers.
Jhun rented an apartment and every time he leaves, he locked it with a
permutation padlock. Unluckily, as Jhun goes back to his apartment he forgot the
4-digit code because of over fatigue. What he remembered from the code are 1, 3, 8,
and 9, the first digit is 1 and the last digit is either 3 or 8.
Guide Questions:
1. How can you help him?
2. What are the possible 4-digit codes?
3. Does the problem relate to combination or permutation? Explain.
Generally, the difference between combinations and permutations is the order
of the elements. That in combinations order does not matter, while in permutations
order matters.
Example: A restaurant offers Do It Yourself salad bar. Vince has chosen fruits like
mango, kiwi grapes, apple, and banana, while Ezekiel chosen grapes, kiwi, banana,
mango and apple.
Do you think their DIY salad are the same?
-As you notice from the problem, both chose the same fruits. Maybe they will differ
from the add-ons they will put into. But if we just consider the fruits, definitely, they
will not.
This implies combinations, which tells us that the arrangement or order of the
fruits is not important.
Problem 1: How many ways can you award a Gold, Silver, Bronze among six teachers
who participated in Hi-Tech ang Guro Ko? Let say the following teachers who
participated were:
A. Mr. Augusto
D. Mr. Dela Vega
B. Mrs. Batas
E. Mrs. Estacio
C. Mrs. Cristobal
F. Mr. Franco
Solution:
Gold Award – There are six choices. The six teachers could be a Gold awardee.
A
B
C
D
E
F
Let say, Mr. Augusto received the Gold Award.
Then, for the Silver Award – There are only five choices. As shown below.
15
DO_Q3_Mathematics10_Lesson4
B
C
D
E
F
Let say, Mrs. Batas is the recipient of Silver Award. Then, there are only four choices
for the Bronze Award.
C
D
E
F
Using Permutations of n objects taken r at a time. Thus,
𝑛!
6!
6•5•4•3•2•1
P(n, r) = (𝑛−𝑟)!
P(6,3) = (6−3)! = 3•2•1 = 6 • 5 • 4 = 120
Therefore, 6x5x4 = 120 ways to award the Gold, Silver and Bronze.
Problem 2: Your teacher asked you to pick three questions in the box with five
questions that you need to answer for your summative quiz. How many different sets
of three questions are there?
Solution:
In this case picking of questions doesn’t matter. For example, you picked 1, 3, and
5, this is not different if you picked 3, 5, and 1. Still, you picked the same numbers.
This implies that this problem can be solve using combinations.
Given: n = 5 and r = 3
𝑛!
5!
5!
5•4•3•2•1
5•4•3!
𝐶(𝑛, 𝑟) = (𝑛−𝑟)!𝑟!
𝐶(𝑛, 𝑟) = (5−3)!3! = 2!3! = 2•1•3•2•1 = 2•3! =10
Thus, there are 10 sets of 3 questions out of 5 questions.
These are 123, 124, 125, 134, 135,145, 234, 235, 245, and 345.
Directions: Solve the following problems. Show your solutions.
1. How many 5 letter codes of 3 consonants and 2 vowels can be formed from 3
consonants and 2 vowels if
a. the first and last letters are both vowel letters?
b. the first and the last letters are both consonants?
c. the middle letter is a vowel?
d. alternately consonant and vowel?
2. Three-digit numbers divisible by 3 are to be formed using any of the digits 0, 1,
3, 5, and 9. How many of this kind can be formed if repetition is not allowed
such that
a. one of the digits is even number?
b. the hundreds digit is even number?
Answer the following problems.
1. Nash has 10 friends, and he wants to invite 6 of them to a party. How many
times will 3 particular friends never attend the party?
2. In a chess competition involving some men and women, every player needs to
play exactly one game with every other player. It was found that in 45 games,
both the players were women and in 190 games, both players were men. What is
the number of games in which one person was a man and other person was a
woman?
16
DO_Q3_Mathematics10_Lesson4
3. In how many different ways can the letters of the word ‘CORPORATION’ be
arranged so that the vowels always come together?
Directions: Write the letter of the correct answer on a separate sheet of paper.
For numbers 1-5, Four different math books, three different economics books, and
five different history books are to be arranged in a shelf. In how many ways can
these books be arranged, if
1. no conditions asked?
A. 12
B. 3! + 4! + 5!
C. 3! 4! 5!
D. 12
2. books of the same kind must be grouped together?
A. 4! 3! 5!
B. 3! + 4! + 5!
C. 4! 3! 5! 3!
D. (3! + 4! + 5!)3!
3. only math books must be grouped together?
A. 4! 8!
B. 4! 9!
C. 4! 8! 9!
D. (4! + 9!)2!
4. only economics books must be grouped together?
A. 3! 9!
B. 3! 10!
C. 3! 4! 5!
D. (3! + 9!)2!
5. only history books must be grouped together?
A. 5! 8!
B. 5! + 8!
C. 5! 7! 8!
D. (5! + 8!)2!
What I Need to Know
This module was designed and written to help students master problems
involving permutations and combinations
After going through this module, you are expected to solve problems involving
permutations and combinations. (M10SP-IIId-e-1).
What I Know
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. If 𝑥 = 𝑃 (4,1), 𝑦 = 𝐶 (3,2), 𝑎𝑛𝑑 𝑧 = 𝐶 (4,4), arrange x, y, and z from smallest to
largest.
A. z, y, x
B. y, z, x
C. x, z, y
D. 𝑥, 𝑦, 𝑧
2. How many 3-digit numbers can be formed from the digits 1,2,3,4,5,6 if no
repetition is allowed?
A. 20
B. 21
C. 100
D. 120
3. In how many ways may the vertices of a quadrilateral be named with the letters
A, B, C, and D?
A. 6
B. 12
C. 24
D. 48
4. If there are 5 doors in a stadium, in how many ways can a person enter one
door and leave by a different door?
A. 5
B. 10
C. 15
D. 20
5. There are 10 players in Tower of Hanoi games. If the elimination round is made
up of a single round robin, how many scheduled matches will there be in the
elimination round?
A. 90
B. 45
C. 25
D. 5
*** If you got an honest score of 5 points (perfect score), you may skip the lesson proper
of each lesson and proceed to the activities or exercises.
17
DO_Q3_Mathematics10_Lesson5
Lesson
5
Solving Problems Involving
Permutations and Combinations
The skills to solve problems allows you to work at your own pace, and you
become more confident to explore and understand the world. Many situations around
us involves permutations and combinations. That’s why, the ideas you have gained
in the previous lesson such as deciding whether the situation is permutation or
combination is really a great help in conquering your fears in solving problems in
real life.
In this lesson, you are expected to solve problems involving permutations and
combinations (M10SP-IIId-e-1).
What’s New
Instead of inviting all your four bosom friends, you have to choose only two
persons to attend to your birthday party. In how many ways you can select 2
persons? Is the order important or not? Explain your answer.
What is it
Recall that in permutation, arrangement or order is very important. While
combination is a selection of things in which order is not important.
1. P (n, r) =
2. P =
𝒏!
(𝒏−𝒓)!
𝒏!
, where 0 ≤ r ≤ n
𝒏𝟏 !𝒏𝟐 !𝒏𝟑 !…𝒏𝒌 !
permutation of n distinct objects taken r at a time.
permutation of identical objects where n is total number of objects and
n1, n2, n3, …, nk are numbers of identical elements
3. 𝑃 = (𝑛 − 1)!
permutation of circular objects
4. 𝐶 (𝒏, 𝒓) =
𝑛!
𝑟! (𝑛−𝑟)!
, where 𝟎 ≤ 𝒓 ≤ 𝒏
combination of 𝒏 distinct objects taken 𝒓 at a time.
For example, choosing only two persons to attend your birthday party among
your four classmates {𝑎𝑛𝑛, 𝑏𝑎𝑏𝑦, 𝑐𝑖𝑛𝑑𝑦, 𝑑𝑒𝑏𝑦}
{a, b}
{a, c}
{a, d}
{b, c}
{b, d}
{c, d}
{b, a}
{c, a}
{d, a}
{c, b}
{d, b}
{d, c}
If order is important, there
are 12 ways in selecting 2
persons among 4 persons.
If 𝑛 = 4 and 𝑟 = 2 using permutation
𝑃 (4, 2) =
4!
(4−2)!
=
4!
2!
=
4(3)(2)(1)
2(1)
= 12 ways
But take note that {a, b} and {b, a} is only one because the list refers to the
same people. Hence, we can reduce the number of ways in selecting two persons
among the 4 choices. Thus, the situation given above is an example of combination.
18
DO_Q3_Mathematics10_Lesson5
{a, b}
{a, c}
{a, d}
{b, c}
{b, d}
{c, d}
{b, a}
{c, a}
{d, a}
{c, b}
{d, b}
{d, c}
There are 6 ways of
selecting 2 persons
among the 4 persons.
Given n = 4 and r = 2 using combination
𝐶 (4,2) =
4!
4!
(4)(3)(2)(1)
=
=
= 6 𝑤𝑎𝑦𝑠
2! (4 − 2)!
2! 2!
(2)(1)(2)(1)
Now, let us deepen your knowledge in solving problems involving permutations and
combinations.
Example 1:
1. In the digits 5, 7, 4, 2, 8, 9 and 3, how many 3-digit numbers can be formed if
repetition of digits is not allowed?
Note that 4! = 4(3)(2)(1)
Solution: 𝑛 = 7 𝑎𝑛𝑑 𝑟 = 3
P (7, 3) =
7!
(7−3)!
=
7!
4!
=
(7)(6)(5)(4!)
4!
= 210 numbers
You may stop expanding the
factorial notation up to 4! to
cancel out 4! in the denominator
Another way of solving this is using the box method approach or the FCP
(Fundamental Counting Principle)
𝑥 6 𝑥 5
7
st
For the 1 box, there are 7 choices, if one of the numbers is chosen, then there
are 6 choices left in the 2nd box and 5 remaining numbers left in the 3rd box.
Thus (7)(6)(5) = 210 numbers can be formed.
2. In how many ways can 4 boys and 3 girls be arranged to sit alternately in a line?
4
𝑥 1 𝑥 1
3 𝑥 3 𝑥 2 𝑥 2
B
G
B
B
G
G
B
Rearranging the factors 4(3)(2)(1)(3)(2)(1) 𝑜𝑟 (4!) (3!) = 144 number of ways
𝑥
𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 4! = 4(3)(2)(1), 𝑎𝑛𝑑 3!
= 3(2)(1)
3. Four children, identical twins, and identical triplets pose for a photograph. How
many photographs can be made?
Solution: n = 9 Getting the sum of 4 children, 2 for twins and 3 from triplets
and n1=2 (identical twins), and n2 = 3 (identical triplets)
9!
𝑃 = 2!3! =
9(8)(7)(6)(5)(4)(3!)
2!3!
= 30, 240 photographs
4. In how many ways can 5 boys and 3 girls be arranged in a circle if the girls must
always stand together?
Solution:
For case 1: 𝑛 = 6 𝐺 𝑏1 𝑏2 𝑏3 𝑏4 𝑏5 , where G is the group of girls. Applying the
formula for circular permutation, we have, 𝑃 = (𝑛 – 1)! = (6 − 1)! = 5!
For case 2: Three girls may arrange themselves. So, there are 3! ways in these
arrangements.
19
DO_Q3_Mathematics10_Lesson5
7
By multiplication rule, we have 𝑃 = 5! 3! = (120) (6) = 720 arrangements
5. How many handshakes will there be if six guests shake hands with all the others
once?
This is an example of combination since the order does not matter.
Solution: 𝑛 = 6 𝑎𝑛𝑑 𝑟 = 2
C (6,2) =
6!
2! (6−2)!
=
6!
(6)(5)(4!)
= (2)(1)(4!)
2! 4!
= 15 ℎ𝑎𝑛𝑑𝑠ℎ𝑎𝑘𝑒𝑠
6. How many triangles can be formed out of points, A, B, C, D, E drawn on the board
of which no three are collinear?
.A
.B
Solution: 𝑛 = 5 𝑎𝑛𝑑 𝑟 = 3
𝐶 (5,3) =
5!
3! (5−3)!
=
5!
3! 2!
(5)(4)(3!)
= (3!)(2)(1) = 10 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠
Note that triangle can be named
using its 3 vertices and the order of
naming them is not important.
.C
.E
.D
7. If 3 balls are picked randomly from a box containing 6 red balls and 8 green balls,
in how many possible ways can it happen that at least 2 of the balls picked are green?
Solution: The order of picking the balls from the box does not matter. Hence,
we are going to use combination.
Case 1: 2 green balls from the box containing 6 red and 8 green balls.
For 𝑛 = 8, 𝑟 = 2 hence, 𝐶 (8,2), but since we need to pick 3 balls
(note that at least 2 balls are green), the additional 1 ball will be from the red
balls. For 𝑛 = 6, 𝑟 = 1, hence, 𝐶 (6,1)
Using multiplication rule, we have 𝑪 (𝟖, 𝟐) 𝒙 𝑪 (𝟔, 𝟏)
Case 2: 3 green balls from the box containing 6 red and 8 green balls.
For 𝑛 = 8, 𝑟 = 3 hence, 𝑪 (𝟖, 𝟑) no need to get red balls from the box
since you need only 3 balls (at least 2 balls are green).
Combining case 1 and case 2, we have: 𝐶 (8,2) 𝑥 𝐶 (6,1) + 𝐶 (8,3) = 224 ways
What’s More
Do as indicated:
A.
1. A circular region is to be covered by 6 different colored sectors of the same
sizes. How many different arrangements are possible?
2. Suppose we have 8 different potted plants, and we wish to arrange them in a
row. If there are only 5 spaces available, in how many ways can this be done?
3. In how many ways can 10 different books be arranged on a shelf?
4. In how many ways can 5 men committee be formed from 8 people?
5. How many ways can the letters of the word “MATHEMATICS” be arranged?
B.
1. How many three-digit even numbers can be formed with the digits 2, 4, 5, 8
and 9 with no repetition allowed?
2. From the word “VIRUS”, how many letter arrangements are possible given
the following conditions:
a. all 5 letters are used without restriction.
20
DO_Q3_Mathematics10_Lesson5
b. all vowels and consonants are together.
3. A box of 10 batteries contains one that is defective. In how many ways can
Leny select 3 batteries such that:
a. the defective battery is included?
b. the defective battery is not included?
4. A jar contains 7 red marbles, 5 white marbles, 3 green marbles. In how
many ways can we select 3 marbles such that:
a. they are all red?
b. They are of different colors?
c. two are red and one is white?
d. exactly one is green?
e. none is green?
What I can
Do
Directions: Answer the following:
1. Solve for the unknown in each item.
a. 𝑃(𝑛, 2) = 110
c. 𝑃(𝑛, 4) = 30𝑃(𝑛, 2)
b. 𝑃 (10, 𝑟) = 720
d. 𝐶(𝑛, 2) = 36
2. How many three-digit numbers can be formed with the digits 1, 2, 3, 4 and 5
if
a. no repetitions allowed
c. the hundreds digit is even number
b. repetition is allowed
d. the ones digit is odd number
Assessment
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. In how many ways can 2 math books, 5 science books, and 3 english books be
arranged in a shelf if they are to occupy just one row?
A. 3 628 800
B. 2 520
C. 30
D. 10
2. How many distinguishable 6-digit numbers can be formed from the digits 343
413
A.1000
B. 720
C. 120
D. 60
3. In how many ways can 30 male teachers and 20 female teachers be elected as
president, vice-president, secretary, and treasurer in their faculty club
association.
A. 𝐶(50,4)
B. 𝑃(50,4)
C. 𝑃 (30, 20)
D. 𝐶(50,20)
4. In how many ways can a clown distribute 9 balloons with different colors to 5
children if each child gets only one balloon?
A.120
B. 126
C. 15 120
D. 362 880
5. How many different pairs of cards can be chosen from the five cards in a royal
flush (A royal flush consists of an ace, king, queen, jack, and 10 of the same suit)
A. 5
B. 10
C. 15
D. 20
This module was designed and written to help you, students, to master
distance formula. The scope of this module permits you to learn over any condition,
especially in this pandemic situation.
21
DO_Q3_Mathematics10_Lesson6
After going through this module, you are expected to illustrate events, and
union and intersection of events (M10SP-lllf-1).
What I Know
Directions: Write the letter of the correct answer in a separate sheet of paper.
1. What is the definition of the union of two mathematical sets, A and B?
A. The
B. The
C. The
D. The
set
set
set
set
of
of
of
of
elements that are in set A, but not in set B.
elements that are in set B, but not in set A.
elements that are in either in set A or set B.
all elements that are both in set A and in set B.
2. If an event is in set R but is not in set S, which of the following statements is
true?
A. The
B. The
C. The
D. The
event
event
event
event
is
is
is
is
in the union of R and S.
not in the union of R and S.
in the intersection of R and S.
in both the union and intersection of R and S.
3. What statement does the shaded region on the right represent?
A. A or B
C. Not A
B. A and B
D. Not B
4.
Refer to the figure on the right, what symbolizes the shaded region?
A. 𝐴 ∪ 𝐵
B. 𝐴
C. 𝐴 ∩ 𝐵
D. 𝐵
5.
Two groups of students, group A and group B surveyed their campus for
student satisfaction. Group A surveyed students from the math, science, and
philosophy departments. Group B surveyed students from the language,
business, and philosophy departments. What could be considered the
universal set in this scenario?
A. The set of all students on campus.
B. The set of all students surveyed by group A.
C. The set of all students surveyed by group B.
D. The set of all students in the philosophy department.
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
Lesson
6
Illustration of Events, Union, and
Intersection of Events
The application of probability specifically in illustrating events and union and
intersection of events can be seen almost everywhere in our daily life such as drawing
an ace from a deck of cards or picking a green piece of candy from a bag of assorted
colors, planning around the weather with what to decide to wear, determining the
best sports strategies to win in the games and competition, analyzing insurance
policies to determine which plans are best for you and your family and what
22
DO_Q3_Mathematics10_Lesson6
deductible amounts you need and many more. You use probability in daily life to
make decisions when you don't know for sure what the outcome will be. The ideas
you have learned in the previous lesson can be used to in illustrating events, and
union and intersection of events.
You own a small canteen near a school, and you offer an affordable combo
meal for students. A combo meal is a combination of a cup of steamed rice, one
serving of veggie, one serving of meat dish, and a free soup.
Every day, you prepare a good number of recipes: two veggies, three meat
dishes, and two types of soup.
As part of your marketing strategy, you give “10+1” promo. That is for every
ten combo meals a group of students buy, one extra combo meal is free. The thrill is
that this extra meal is given randomly drawing a slip of paper from a box, wherein
the combo meal is written.
Guide Questions:
1. What are the possible combinations for this extra combo meal?
2. How many such combinations?
Any activity that involves chance such as rolling a die is an experiment. The
result of an experiment is an outcome. The set of all different outcomes of an
experiment is called sample space of the experiment. In describing the sample space
of an experiment, you will write all the possible outcomes of the experiment. You
usually use small letters to denote outcomes and the capital letter 𝑆 to denote the
sample space. The total number of outcomes in a sample space is denoted by the
symbol 𝑛(𝑆). An event is a set of one or more outcomes.
Example 1: An experiment is performed by rolling a single die. The sample is
S={1,2,3,4,5,6}
Let A = event “an even number is rolled”
B = event “a number greater than two is rolled”
Illustrate the events in the experiment using Venn Diagram.
a. What is A? B?
b. What is 𝐴 ∪ 𝐵 ?
c. What is 𝐴 ∩ 𝐵 ?
Solutions:
a. Illustration of events in the experiment using Venn Diagram.
A= {2,4,6}
B= {3,4,5,6}
b. A union B denoted as 𝑨 ∪ 𝑩 = {2,3,4,5,6}
23
DO_Q3_Mathematics10_Lesson6
c. A intersection B denoted as 𝑨 ∩ 𝑩 ={4,6}
Example 2: In rolling a pair of dice (one white and one black), we can have the
following events:
A= {(1,6)}
B= {(1,6), (6,1)}
C= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)}
D= {(1,1), (1,6), (6,1), (6,6)}
E= {(1,1), (1,3), (5,5)}
a. Which events are compound events?
d. What is 𝐴 ∩ 𝐵 ?
b. Which events are subsets of the other?
e. What is 𝐵 ∩ 𝐷?
c. What is 𝐴 ∪ 𝐵 ?
f. What is 𝐷 ∪ 𝐸?
Solutions:
a. All events except event A are compound events since each of the four events B,
C, D and E has 2,6,4 and 3 outcomes, respectively.
b. 𝐴 ⊂ 𝐵 since the outcome (1,6) is also in B. 𝐵 ⊂ 𝐷 since the outcomes (1,6) and
(6,1) are also in D.
c. 𝐴 ∪ 𝐵 = {(1,6), (6,1)}
d. 𝐴 ∩ 𝐵 = {(1,6)}
e. 𝐵 ∩ 𝐷 = {(1,6), (6,1)}
f. 𝐷 ∪ 𝐸 = {(1,1), (1,3), (1,6), (5,5), (6,1), (6,6)}
Example 3: An experiment is performed by simultaneously spinning the two
spinners, the first spinner containing the vowels in the alphabet and the second
having the first five counting numbers. Let x be the event that a spinner stops at an
even number and y be the events that a spinner stops at a consonant letter. Find
𝑥 ∪ 𝑦 and 𝑥 ∩ 𝑦.
Solutions:
Illustration of events in the experiment by using Venn Diagram
x = {2A,4A,2E,4E,2I,4I,2O,4O,2U,4U}
y = {}
𝒙 ∪ 𝒚 = {2A,4A,2E,4E,2I,4I,2O,4O,2U,4U} 𝒙 ∩ 𝒚 = { } or called as impossible set
since there is no consonant letters present in the spinner.
Answer the following given the problems.
A. A two-child family is selected at random. Let B denote the event that at least one
child is a boy B = {𝑏𝑏, 𝑏𝑔, 𝑔𝑏}, let D denote the event that the genders of the two
children differ D= {𝑏𝑔, 𝑔𝑏}, and let M denote the event that the genders of the two
children match M ={𝑏𝑏, 𝑔𝑔}.
Find:
1. B ∪ D: _______________
4. B ∩ D: ___________________
24
DO_Q3_Mathematics10_Lesson6
2.
3.
D ∪ M: _______________
B ∩ M:_______________
5. B ∪ D ∪ M : _______________
Use the Venn Diagram to answer the following questions.
In a class of 60 students, 21 sign up for chorus, 29 sign up for band, and 5
take both, 15 students in the class are not enrolled in either band or chorus.
1. Illustrate the given situation in the Venn Diagram.
2. How many are in set A?
3. How many are in set B?
4. What is 𝐴 ∪ 𝐵 ?
5. What is 𝐴 ∩ 𝐵 ?
Directions: Write the letter of the correct answer in a separate sheet of paper.
For item #s 1-5. Please refer to the figure below.
J
S
T
1. What is 𝑱 ∩ 𝑺 ∩ 𝑻 ?
A. 20
B. 25
C. 29
D. 36
B. 4
C. 9
D. 36
B. 57
C. 62
D. 63
C. 5
D. 9
C.9
D. 16
2. Find 𝑱 ∩ 𝑺 ?
A. 3
3. What is 𝑺 ∩ 𝑻 ?
A. 29
4. How many can dance jive only?
A. 3
B.4
5. How many can dance salsa only?
A. 4
B. 5
25
DO_Q3_Mathematics10_Lesson6
This module was designed and written to help you, students, to master
probability of a union of two events. The scope of this module permits you to learn
over any condition, especially in this pandemic situation. The language used in this
module recognizes various vocabulary levels of the students. The lessons are
arranged based on the standard sequence of the Most Essential Learning
Competencies (MELCs) of the Basic Education Learning Continuity Plan (Be-LCP).
After going through this module, you are expected to illustrate the probability
of a union of two events (M10GE-IIIg-1).
What I Know
Direction: Write the letter of the correct answer in a separate sheet of paper.
1. A card is drawn from a standard deck of cards. Find the probability of drawing
a diamond or a 5?
A.
4
13
B.
5
13
C.
9
26
D.
17
52
Given for items 2-4: Two fair dice are rolled. If A = {both numbers are even},
B = {one number is 3 and the other is even} and C = {a sum less than 5 appears}
2. Which set contains the elements of A?
A. {2,4,6,8}
B. {(2,2), (2,4), (2,6), (4,6), (4,4), (6,6)}
C. {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,6)}
D. {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4)(6,6)}
3. What is 𝐴 ∪ 𝐵?
A. { }
B. {(3,2), (3,4), (3,6)(2,3), (4,3), (6,3)}
C. {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4)(6,6)}
D. {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4)(6,6), (3,2), (3,4), (3,6)(2,3), (4,3), (6,3)}
4. What is the probability of 𝐴 ∪ 𝐵?
A. 0
B.
1
12
C.
1
6
D.
5
12
5. Anna has 70% chance to pass her Grade 10 English, 78% to pass her Grade 10
Mathematics and 60% chance to pass her both subject. What is the probability
that she will pass at least one of these subjects?
A. 75%
B. 78%
C. 88%
D. 95%
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
26
DO_Q3_Mathematics10_Lesson7
Lesson
7
Probability of a Union of Two Events
All statistics classes include questions about probabilities involving the union
of sets. In English, we use the word "and" to describe this concept. For example,
"Find the probability that a college student is taking a mathematics class or a science
class." That is expressing the union of the two sets in words. In this lesson, you will
learn how to find the probability of a union of two events.
A tutoring service specializes in preparing adults for college entrance tests.
The Venn Diagram below shows who among the students ask for help from the
tutoring service. Find the probability that students need help in:
1. Mathematics
2. English
3. both subjects
4. either Mathematics or English
The union of events 𝐴 and 𝐵, denoted 𝐴 ∪ 𝐵, is the collection of all outcomes
that are elements of one or the other of the sets 𝐴 and 𝐵, or of both. It corresponds
to combining descriptions of the two events using the word “or”. To say that the event
𝐴 ∪ 𝐵 occurs means that on a particular trial of experiment either 𝐴 or 𝐵 occurred (or
both did)
To find the union of two sets, list the elements that are in either (or both) sets.
In terms of a Venn Diagram, the union of sets 𝐴 and 𝐵 can be shown as two
completely shaded interlocking circles.
In symbols, since the union of A and B contains all
the points that are in A or B or both, the definition of the
union is: 𝐴 ∪ 𝐵 = {𝑥|𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}.
For example, if 𝐴 = {2,4,6,8,10} and 𝐵 = {5,10,15} , then 𝐴 ∪ 𝐵 = {2,4,5,6,8,10,15}.
Notice that the element 10 is not listed twice in the union, even though it appears in
both set A and B. This leads us to the general addition rule for the union of two
events: 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) where 𝑃(𝐴 ∩ 𝐵) is the intersection of the
two sets. It contains all the outcomes that is common to both events, and it is the
intersection of the two circles in the Venn diagram. We must subtract this out to
avoid double counting of the inclusion of an element.
27
DO_Q3_Mathematics10_Lesson7
If sets 𝐴 and 𝐵 are disjoint, however, the event 𝐴 ∩ 𝐵 has no outcomes in it,
and is an empty set denoted as ∅, which has a probability of zero. So, the above rule
can be shortened for disjoint sets only:
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
Example 1:
Suppose that in your class of 30 students, 8 students love to sing, 15 students
like to dance, and 5 students are both love to sing and dance. Find the probability
that a student loves to sing or dance.
Let A be the event that a student loves to sing and let B be the event that a
student likes to dance. Let us create a Venn diagram that models this situation.
𝑷(𝑨 ∪ 𝑩) =
3 + 5 + 10
30
18
= 30
[Add]
3
[Simplify]
𝑷(𝑨 ∪ 𝑩) = 5
You could also compute this probability using the Addition Rule:
𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩)
=
=
=
=
8
15
5
+ 30 − 30
30
[Substitute]
8+15−5
30
23−5
30
[Combine terms]
[Add]
18
30
[Subtract]
3
[Simplify]
𝑷(𝑨 ∪ 𝑩) = 5
𝟑
Therefore, the probability that a student loves to sing or dance is .
𝟓
Example 2:
In rolling a single die, find the probability of getting an odd number or a
number greater than 5.
A die is a cube with six faces and each side marked with a different number
of dots from one to six. The outcomes of getting an odd number is 𝐸 = {1,3,5} and a
number greater than 5 is 𝐺 = {6}. Since they have no elements in common it means
that it is impossible for both events to occur on a single trial of the random
experiment.
3+1
𝑷(𝑬 ∪ 𝑮) =
6
=
4
6
[Add]
2
[Simplify]
𝑷(𝑬 ∪ 𝑮) = 3
Therefore, the probability of getting an odd number or a number greater
𝟐
than 5 when rolling a single die is 𝟑.
28
DO_Q3_Mathematics10_Lesson7
A. A card is drawn at random from a standard deck of 52 playing card. Find the
probability of each outcome.
1. A red card or a king
2. An ace or a number card
3. A jack or a diamond
4. A black card or a spade
5. A face card or a number 5
B. A spinner is subdivided into 8 equal parts with numbers 1-8 written on it. The
spinner may stop on any of the eight numbered sectors of the circle. (Assume
that the spinner will not stop on the line between two sectors.) Find the
probability of each outcome.
1. An odd number or a number divisible by two.
2. A prime number or a six.
3. A number less than 5 or an even number.
4. A seven or a number greater than 4.
5. A number greater than six or a composite number.
The number of students enrolled in Algebra Class (A) and Geometry
class(G) in a certain university is represented by Venn diagram. Study the
figure below and answer the questions that follow.
a. What is the probability that a student is
enrolled in the Algebra class?
b. What is the probability that a student is
enrolled in a Geometry class?
c. What is the probability that a student is both
enrolled in a Geometry and Algebra class?
d. What is the probability that a student is
enrolled either in a Geometry or Algebra class?
Directions: Write the letter of the correct answer in a separate sheet of paper.
1. Events A and B are defined by the sets {1,2,3,4} and {3,4,5,6}, respectively. What is
the union of two events?
A. {1,2,3, 3,4,4,5,6}
B. {1,2, 3,4,5,6}
C. {1,2,5,6}
D. {3,4}
Given for items 2-5: Raul rolled a fair die and finds the probability of the number
that turns up even or divisible by 4.
2. Which of the following condition is true about its probability?
A. The probability is less than 1.
B. The probability is greater than 0.
C. The probability is greater than 0 but less than or equal to 1
D. The probability is greater than or equal to 0 but less than or equal to 1.
29
DO_Q3_Mathematics10_Lesson7
3. Which formula to be used to find its probability, if 𝑃(𝐴) is the probability of turning
up an even number and P(B) is the probability of turning up a number divisible
by 4?
A. 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∪ 𝐵)
C. 𝑃(𝐴) + 𝑃(𝐵)
B. 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
D. 𝑃(𝐴) ∙ 𝑃(𝐵)
4. What is the probability that a number turns up an even number or divisible by 4?
1
1
1
2
A. 2
B. 3
C. 6
D. 3
5. A card is picked at random from well shuffled standard playing cards. What is the
probability that the card picked is a face card or a heart?
25
11
1
2
A. 52
B. 26
C. 2
D. 3
This module was designed and written to help you, students, to master the
topic finding the probability of (A U B). The scope of this module permits you to learn
over any condition, especially in this pandemic situation. The language used in this
module recognizes various vocabulary levels of the students. The lessons are
arranged based on the standard sequence of the Most Essential Learning
Competencies (MELCs) of the Basic Education Learning Continuity Plan (Be-LCP).
After going through this module, you are expected to find the probability of
(𝐴 𝑈 𝐵) (M10SP-lllh-1).
What I Know
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. In a game of chess, the die decides who go first. Both players roll the die and
whoever gets the higher number plays first. Player A gets a 2. What is the
probability that player B gets to play first?
1
2
1
3
A.
B.
C.
D.
3
3
4
4
2. A die is rolled. What is the probability of getting an even number or a factor of 2?
1
2
3
4
A. 2
B. 3
C. 4
D. 5
For numbers 3-4: consider this problem: A cube with A, B, C, D, E, and F on its faces
is rolled.
3. What is the probability of rolling a vowel or a letter in the word FRAUD?
2
1
3
A. 3
B. 2
C. 4
D. 1
4. What is the probability of rolling a consonant or a vowel?
A.
2
3
B.
1
2
C.
3
4
D. 1
5. A box of miniature cars contain 6 red cars, 5 blue cars and 7 black cars. One car
is drawn at random. Find the probability of getting either red or blue.
1
7
11
2
A. 3
B. 18
C. 18
D. 3
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
30
DO_Q3_Mathematics10_Lesson8
Lesson
8
Finding the Probability of Union and
Intersection of Events
Not all events are simple, some would require a combination of these simple
events that would turn it into a compound event. In real life, some events are
connected and so we must take into consideration every detail present in each
circumstance. In the last module, you are familiarized with the concept of the
probability of the union of two events. This module would deepen your knowledge
and skills in dealing with this type of probability.
Among siblings, it really is difficult to know who would do a certain household
chore. Usually, the “bunso” would always do these tasks. To prevent this instance,
Santos Family decided to use a die to know who gets the right to do certain chores.
Whoever gets the higher number on the die would do the task. On a holiday, Uno
and Dos decided to use this trick to know who would go to Puregold Paso de blas to
get some groceries. Uno rolls a 4.
Guide Questions:
1. What can you say about the event in this problem?
2. What number/s Dos needs to do the chore?
3. How can we find the probability in this situation?
The above scenario is an example of a compound event. From the situation
presented, Dos will do the chore if he gets a higher number on the die. Since Uno
rolls a 4, Dos needs to get a 5 or a 6 to do the chore. The P(5) =
1
6
1
6
while the P(6)= .
For us to find the probability of getting a 5 or a 6, or that is the probability that Dos
would do the chore, we must add the corresponding probability of getting a 5 and
getting a 6. So we have,
1
6
1
+6 =
2
6
1
𝑜𝑟 3.
Generally, we find the probability of the union of events by adding the
corresponding probabilities. But, if there is a common element, we must subtract it
with the initial sum so that we will not count it twice. In symbols, we have:
𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) ; if there is/are no common element/s
𝑃 (𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) – 𝑃(𝐴 ∩ 𝐵) ; if there is a common element
Example 1. A box contains 4 red balls, 5 green balls, and 3 blue balls. Helen
draws one ball at random. What is the probability that the ball is either green
or red?
In this problem, a certain ball cannot be both red and green, that is why there
are no common elements. Thus, we need to apply the first formula, 𝑃(𝐴 𝑜𝑟 𝐵) =
𝑃(𝐴) + 𝑃(𝐵). We are looking for the probability that it is either green or red. So:
5
4
𝑃(𝐺𝑟𝑒𝑒𝑛 𝑜𝑟 𝑅𝑒𝑑) = 𝑃(𝐺𝑟𝑒𝑒𝑛) + 𝑃(𝑅𝑒𝑑) 𝑃(𝐺𝑟𝑒𝑒𝑛 𝑜𝑟 𝑅𝑒𝑑) =
+
12 12
𝑃(𝐺𝑟𝑒𝑒𝑛 𝑜𝑟 𝑅𝑒𝑑) =
31
9
3
𝑜𝑟
12
4
DO_Q3_Mathematics10_Lesson8
Example 2. Card grades ranges from 70 to 100. What is the probability of
getting a grade which is a multiple of 3 or a multiple of 2?
In this problem, let us try to list the elements of the given conditions. Let A=
{multiples of 3 from 70 to 100} and B= {multiples of 2 from 70 to 100}
Set A = {72,75,78,81,84,87,90,93,96,99} ; n(A)=10
Set B = {70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,100} ; n(B)=16
As you can see, there are common elements, these are 72,78,84,90,96. We
denote this as 𝐴 ∩ 𝐵. We now use the second formula since there are common
elements (5 common elements).
𝑃 (𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)– 𝑃(𝐴 ∩ 𝐵)
𝑃(𝐴 𝑜𝑟 𝐵) =
10
16
5
+
–
31
31 31
𝑃(𝐴 𝑜𝑟 𝐵) =
21
31
Example 3. Two numbers are chosen simultaneously from 1,2, 3, …, 9 at
random. Find the probability that their sum is even.
We are asked to find the probability that the sum is even if we pick two
numbers from 1 to 9 at random. There can be two cases that our sum can be even:
(1) both numbers are odd; (2) both numbers are even.
We let 𝐴 = the case when both numbers are odd and
𝐵 = the case when both numbers are even.
For Set A, we have the elements 1,3,5,7,9 and we are to pick 2, so we have
5C2. For set B, we have the elements 2,4,6,8 and we are also to pick 2, so we have
4C2. For our sample space, we have 9C2. Therefore:
5 𝐶2
4 𝐶2
𝑃(𝑠𝑢𝑚 𝑖𝑠 𝑒𝑣𝑒𝑛) =
+
9 𝐶2
9 𝐶2
𝑃(𝑠𝑢𝑚 𝑖𝑠 𝑒𝑣𝑒𝑛) =
10
6
+ 36
36
𝑃(𝑠𝑢𝑚 𝑖𝑠 𝑒𝑣𝑒𝑛) =
16
36
𝑜𝑟
4
9
Find the probability of each event happening.
1. A card is drawn at random from a standard deck of cards. What is the
probability of drawing a queen or a king?
2. A die is rolled. What is the probability of getting a prime number or an odd
number?
3. A spinner is divided into 6 equal parts numbered 1-6. If it is spun twice, find
the probability that the sum of the first spin and the second spin is more than
8.
4. A pouch contains 10 candies of the same shape and size. Five of the candies
are mints, 4 are toffees and 1 is chocolate. If one candy is picked at random,
what is the probability that it is a mint or a chocolate?
5. A scrabble has 100 tiles. Nine of them are A’s, 12 are E’s, 9 are I’s, 8 are O’s
and 4 are U’s. What is the probability that a randomly selected scrabble tile is
a vowel?
32
DO_Q3_Mathematics10_Lesson8
Solve the following problems.
1. If there is 30% chance of rain on Saturday, 70% chance of rain on Sunday,
and 21% chance of rain on Saturday and Sunday, what is the probability that
it will rain on either Saturday or Sunday?
2. The letters of the word VALENZUELA and MODULE are written on individual
cards and the cards are placed in a bag. If a card is picked at random from the
bag, what is the probability of picking a vowel?
3. On New Year’s Eve, the probability of the person having a car accident is 0.07.
The probability of the person driving while intoxicated is 0.3, and the
probability of a person having a car accident while intoxicated is 0.15. What is
the probability of a person driving while intoxicated or having a car accident?
Directions: Write the letter of the correct answer on a separate sheet of paper.
For numbers 1-2, consider this problem: A box of miniature cars contain 6 red
cars, 5 blue cars and 7 black cars. One car is drawn at random.
1. Find the probability of getting either blue or black.
1
7
11
2
A. 3
B. 18
C. 18
D. 3
2. Find the probability of getting neither blue nor black.
1
7
11
2
A. 3
B. 18
C. 18
D. 3
For numbers -5, consider this problem: In a race for barangay captain, 4
candidates Q, R, S, and T have 0.14, 0.18, 0.25, and 0.40 probabilities of winning,
respectively.
3. What is the probability that Q, R, S, or T will win?
A. 0.32
B. 0.54
C. 0.57
D. 0.97
4. What is the probability that either Q or T will win?
A. 0.32
B. 0.54
C. 0.57
D. 0.97
5. The probability of 3 teams C, M and N of winning the tournament are
2 1
1
, 𝑎𝑛𝑑 12, respectively. Assuming that only 1 team wins the tournament,
9 6
find the probability that either M or N win.
1
4
7
11
A. 4
B. 7
C. 18
D. 36
This module was designed and written to help you, students, to identify
situations that illustrate mutually exclusive events and master solving problems
involving probability. The scope of this module permits you to learn over any
condition, especially in this pandemic situation.
After going through this module, you are expected to illustrate mutually
exclusive events (M10SP-IIIi-1) and solve problems involving probability(M10SP-IIIij-1).
33
DO_Q3_Mathematics10_Lesson9
What I Know
Direction: Write the letter of the correct answer on a separate sheet of paper.
1. If a family has three children, what is the probability that all are boys?
1
3
1
1
A.
B.
C.
D.
3
8
8
6
2. A class consist of 5 Bicolanos, 4 Tagalogs, 3 Ilonggos and 8 Ilocanos. A
student is chosen at random to represent the class. Find the probability that
the student is an Ilocano or an Ilonggo?
3
8
9
11
A. 20
B. 20
C. 20
D. 20
Given for items 3 -4: A purse contains nine 5-peso coins and seven 10-peso coins.
If 3 coins are drawn at random.
3.
What is the probability that all are 5- peso coins?
3
4
3
9
A. 20
B. 5
C. 16
D. 16
4. What is the probability that one is a 5- peso coin and two are 10- peso
coins?
3
27
5
9
A. 20
B. 80
C. 16
D. 16
5. A card is picked at random from well shuffled standard playing cards. What
is the probability that the card picked is a face card or a heart?
25
11
1
2
A.
B.
C.
D.
52
26
2
3
*** If you got an honest score of 5 points (perfect score), you may skip the lesson
proper of each lesson and proceed to the activities or exercises.
Lesson
9
Mutually Exclusive Events and Solving
Problems Involving Probability
If you win in a competition, can you be declared as the loser of the game? In
riding a bicycle, can you turn left and right at the same time? There are just some
events which cannot happen simultaneously or at the same time. In probability, we
call it mutually exclusive events. In this lesson, you will learn about mutually
exclusive events and solve problems involving probability as well.
In the experiment of tossing a coin, either heads “H’ or tails “T” might show
up, but not both heads and tails at the same time. Similarly, in throwing a die only
one number can show up on the top face.
1. In finding the probability of each event above, what concepts are needed?
2. How will you compute for the probability in each case?
34
DO_Q3_Mathematics10_Lesson9
Two events, A and B, are disjoint or mutually exclusive if and only if the
probability of events A and B occurring together is zero. This means that they cannot
happen at the same time. That is,
𝑃(𝐴 ∩ 𝐵) = 0
below.
In other words, events A and B cannot overlap as shown in the Venn diagram
B
For example, if 𝐴 = {1,2,3} , 𝐵 = {5,6} then 𝐴 ∩ 𝐵 = { }. The sets A and B have
no elements in common which means the sets are disjoint thus, their intersection is
an empty set. Furthermore, the probability of the disjoint set, A and B, is zero.
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝑨 ∩ 𝑩) = 𝟎
Example 1:
Recall that in a standard deck of cards, there are 4 suites: heart, clubs, diamond,
and spade. Each suite contains 13 cards (10 cards from ace to 10 and 3 face cards jack, queen, and king).
Clubs (Black)
Spade (Black)
Heart (Red)
Diamond (Red)
Only the heart and diamond suites are color red while the clubs and spade are
black (explanation is needed as the image shown above may be printed in black
and white only).
Now, Compute for the probability of:
a. Drawing a heart (event H) and a black card (event B)?
35
DO_Q3_Mathematics10_Lesson9
In a standard deck of cards only the heart and diamond suites are red,
which means it is impossible for us to draw a heart card that is also a black
card in one draw. Using Venn diagram, we can see that the two sets do not
overlap which means they do not have any elements in common. This means
that the two sets H and B are mutually exclusive.
Thus, the probability of drawing a
heart and a black card is zero.
𝑷(𝑯 𝒂𝒏𝒅 𝑩) = 𝑷(𝑨 ∩ 𝑩) = 𝟎
b. Drawing a heart (event H) or a black
card (event B)?
The probability of drawing a heart OR a black card is possible simply by
adding the probability of each occurring. Thus,
𝑷(𝑯 𝒐𝒓 𝑩) = 𝑷(𝑯 ∪ 𝑩) = 𝑷(𝑯) + 𝑷(𝑩)
𝟏𝟑
𝟐𝟔
= 𝟓𝟐 + 𝟓𝟐
𝟑𝟗 𝟑
= = 𝟎. 𝟕𝟓 𝒐𝒓 𝟕𝟓%
𝟓𝟐 𝟒
𝑷(𝑯 𝒐𝒓 𝑩) =
Thus, the probability of drawing either a heart or a black card is 75%.
Solving Problems Involving Probability
Along with other topics we have already discussed in your previous module,
let us solve word problems involving probability.
Example 2:
In a group of 40 people,10 are healthy and every person of the remaining 30
has either high blood pressure, a high level of cholesterol or both. If 15 have high
blood pressure and 25 have high level of cholesterol:
a. how many people have high blood pressure and a high level of cholesterol?
Solution:
Let x be the number of people with both high blood pressure and high level
of cholesterol.
Hence, (15 - x) will be the number of people with high blood pressure ONLY,
and (25 - x) will be the number of people with high level of cholesterol ONLY.
We now express the fact that the total number of people with high blood
pressure only, with high level of cholesterol only and with both is equal to
30. Thus, (15 - x) + (25 - x) + x = 30. Using the above equation, we can now
solve for the number of people who have high blood pressure and high level
of cholesterol which we represented as x.
(15 - x) + (25 - x) + x = 30
40 – x = 30
Simplify by combining similar terms
40 – 30 = 𝑥
Use addition property of equality to solve for x
10 = 𝑥 or 𝑥 = 10
36
DO_Q3_Mathematics10_Lesson9
Thus, 10 people have both high blood pressure and high level of
cholesterol.
b. If a person is selected randomly from this group, what is the probability that
he/she has high blood pressure (event A)?
From the problem, 15 out of 40 people have high blood pressure. Therefore,
the probability of event A occurring is:
𝑃(𝐴) =
15
= 0.375 = 37.5%
40
Thus, when selected at random, the probability that a person has high
blood pressure is 0.375 or 37.5%
a. has high level of cholesterol (event B)?
From the problem, 25 out of 40 people have high level of cholesterol.
Therefore, the probability of event B occurring is:
𝑃(𝐵) =
25
= 0.625 = 62.5%
40
Thus, when selected at random, the probability that a person has high
level of cholesterol is 0.625 or 62.5%.
A. Solve the following problems.
1. There are 200 birds in a zoo. 70 birds are male with brown eyes and 100
birds are female with brown eyes. 20 of the birds are male with blue eyes
and 10 birds are female with blue eyes. If a bird us selected at random,
what it the probability that the bird is:
(Hind: you can create a table to further help you answer the questions, just
like in question A)
a. a female?
b. a male with brown eyes?
c. a female given that it has brown eyes?
d. a male given that it has blue eyes?
e. Creature with blue eyes given that it is a female?
B. Solve the following problems.
1. What is the probability that a person is born in a month that ends in ‘er’?
2. If a number is randomly chosen from the list: 32, 49, 55, 30, 56, 28, 50,
40, 40, 45, 3, 25 what is the probability that the number is a multiple of
5?
3. In a donut shop, 63% of customers buy a donut, 80% of customers buy
coffee, and 88% buy coffee or a donut. What percentage of customers buy
coffee AND a donut?
4. A special deck of 16 cards has 4 that are blue, 4 yellow, 4 green and 4 red.
The four cards of each color are numbered from one to four. A single card
is drawn at random. Find the probability of drawing a card that is red or
the number on the card is at most 3?
37
DO_Q3_Mathematics10_Lesson9
Solve the following problems. Write your answers in your paper.
A. In a class in which all students practice at least one sport, 60% of students
play soccer or basketball and 10% practice both sports. If there is also 60%
that do not play soccer, calculate the probability that a student chosen at
random from the class:
1. Plays soccer only.
2. Plays only one of the sports.
3. Plays neither soccer nor basketball.
B. In a classroom there are 100 pupils, of whom 40 are boys, 30 wear glasses,
and 15 are boys who wear glasses. If one student from the class is randomly
selected:
1. What is the probability that the student will be a girl who does not wear
glasses?
2. If we know that the student selected does not wear glasses, what
probability that it will be a boy?
Directions: Write the letter of the correct answer on a separate sheet of paper.
1. There are 10 students in a class including Rannie and Grace. If two students
are to be chosen at random as class representatives, find the probability that
both Rannie and Grace are chosen?
1
1
4
44
A.
B.
C.
D.
25
45
2. The probability that Paul will hit a target is
lose is
2
.
3
45
1
and
3
45
the probability that he will
What is the probability that in three shoots, he will hit the target any
two times and lose one time?
A.
1
2
B.
1
3
C.
1
6
D.
2
9
Given for items 3-5: First 50 counting numbers are written on slips of paper,
placed in a box, and mixed thoroughly. Two numbers are drawn together at
random.
3. Find the probability that both numbers drawn are odd.
A.
3
49
B.
6
49
C.
9
49
D.
12
49
D.
12
49
4. Find the probability that both numbers are even.
A.
3
49
B.
6
49
C.
9
49
5. Find the probability that the number selected is an even or a multiple of 4
A.
3
49
B.
6
49
C.
9
49
D.
12
49
For inquiries or feedback, please write or call:
Department of Education – SDO Valenzuela
Office Address:
Pio Valenzuela Street, Marulas, Valenzuela City
Telefax:
(02) 8292-4340
Email Address:
sdovalenzuela@deped.gov.ph
38