Permutations – it is an arrangement of a group of things in a definite order, that is, there is first element, a second,
a third, etc. In other words, the order or arrangement of the elements is important.
Combination – also concerns arrangements, but without regard to the order. This means that the order or
arrangement in which the elements are taken is not important.
Difference between permutation and combination:
Example Problem: Determine the number of possible arrangements of two letters from a set of four (ABCD), using
principles from permutations and combinations.
Permutation (order or arrangement is important):
Solution #1:
AB
AC
AD
BA
BC
BD
CA
CB
CD
DA
DB
DC
Therefore, we can arrange 12 two-letters from the given set of four letters.
Solution #2:
Using permutation formula: Arrangement of “n” object taken “r” at a time.
𝑛𝑃𝑟 =
𝑛!
(𝑛 − 𝑟)!
n=4
r=2
4𝑃2 =
4!
4! 4𝑋3𝑋2𝑋1
= =
= 𝟏𝟐
(4 − 2)! 2!
2𝑋1
Solution #3:
Using Calculator Technique:
Press 4 → Shift + x → 2 → = (the answer will be 12)
Combination (without regard to the order):
Solution #1:
AB
AC
AD
BA
BC
BD
CA
CB
CD
DA
DB
DC
We only have six arrangements; having the same letters in the arrangements but different positions are still the
same arrangements.
Solution #2:
Using combination formula:
𝑛𝐶𝑟 =
𝑛𝑃𝑟
𝑛!
=
𝑟!
𝑟! (𝑛 − 𝑟)!
n=4
r=2
𝑛𝐶𝑟 =
𝑛!
4!
4!
4𝑋3𝑋2! 4𝑋3 12
=
=
=
=
=
=𝟔
𝑟! (𝑛 − 𝑟)! 2! (4 − 2)! 2! (2!) 2! (1𝑋2)
2
2
Solution #3:
Using Calculator Technique:
Press 4 → Shift + ÷ → 2 → = (the answer will be 6)
A. Permutations
Types of Permutations:
1. Arrangement of “n” object taken “n” at a time.
- The number of permutations of n distinct object is n!
Formula:
nPn = n! = n(n-1) (n-2) (n-3) …(n-k)
Example:
1a. In how many ways can 5 Engineering Board Examinees be lined up to go inside the testing centers?
5P5 = 5! = 5(5-1) (5-2) (5-3) (5-4) = 5 (4) (3) (2) (1) = 120 ways
1b. In how many ways can four dating reviewees be seated in the review center without restrictions?
8P8 = 8! = 8X7X6X5X4X3X2X1 = 40,320 ways
2. Arrangement of “n” object taken “r” at a time.
- The number of permutations of n distinct object taken r at a time. It is the type of permutation of
determining the number of arrangements by taking a sample (r) out of a given population.
Formula:
𝑛!
𝑛𝑃𝑟 =
(𝑛 − 𝑟)!
Example:
2a. In “Tawag ng Tanghalan” contest, the 10 finalists consist of 4 males and 6 females. Find the number of sample
points for the number of possible orders at the conclusion of the contest for the first three positions.
n = 10
r=3
10𝑃3 =
10!
10𝑋9𝑋8𝑋7!
=
= 10𝑋9𝑋8 = 𝟕𝟐𝟎 𝒘𝒂𝒚𝒔
(10 − 3)!
7!
2b. A basketball team consists of 14 players at a time. In how many ways can 5 starting positions be filled if all of
them can play any of the positions?
n = 14
r=5
14𝑃5 =
14!
14𝑋13𝑋12𝑋11𝑋10𝑋9!
=
= 14𝑋13𝑋12𝑋11𝑋10 = 𝟐𝟒𝟎, 𝟐𝟒𝟎 𝒘𝒂𝒚𝒔
(14 − 5)!
9!
3. Arrangement of “n” object some are alike.
- The number of distinct permutations of n things of which n1 is of one kind, n2 of a second kind,….., nk
of a kth kind is
Formula:
𝑛!
(𝑛1 !)𝑥((𝑛2 !)𝑥(𝑛3 !)𝑥 … (𝑛𝑘 !)
Example:
3a. A school plays 10 basketball games in the SCUAA per season. In how many ways can the CSPC Blue Stallions
end their season with 6 wins and 4 losses?
n = 10
𝑛1 = 6
𝑛2 = 4
10!
10𝑋9𝑋8𝑋7𝑋6! 10𝑋9𝑋8𝑋7 5040
=
=
=
= 𝟐𝟏𝟎 𝒘𝒂𝒚𝒔
(6!)(4!) 6! (4𝑋3𝑋2𝑋1)
4𝑋3𝑋2𝑋1
24
4. Circular Permutations
- Permutation that occurs by arranging objects in a circle. To find the number of ways of arranging n
different objects in a circle, we first fix or select a position for one of the objects.
Formula:
(𝑛 − 1)!
Example:
4a. In how many ways can 6 students be seated in a round dining table?
n=6
(6 − 1)! = 5! = 5𝑋4𝑋3𝑋2𝑋1 = 𝟏𝟐𝟎 𝒘𝒂𝒚𝒔
4b. In how many ways can a caravan of 8 covered Kalesa from Intramuros be arranged in an oval track?
n=8
(8 − 1)! = 7! = 7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1 = 𝟓, 𝟎𝟒𝟎 𝒘𝒂𝒚𝒔
B. Combinations
Combinatorial Formula
-
The number of ways of selecting r objects take from n at a time is C (n, r) or nCr where
𝑛𝑃𝑟
𝑛!
𝑛𝐶𝑟 =
=
𝑟!
𝑟! (𝑛 − 𝑟)!
Example
B1. How many ways can three people from Dave, Dan, Julie, Amanda, and Steve be selected to attend a meeting?
n=5
r=3
5𝐶3 =
5!
5𝑋4𝑋3!
5𝑋4 20
=
=
=
= 𝟏𝟎 𝒘𝒂𝒚𝒔
3! (5 − 3)! (3!)(2𝑋1) 2𝑋1
2
B2. How many ways are there to select 3 candidates for a production planner position from 15 equally qualified
IE applicants?
n = 15
r=3
15𝐶3 ==
15!
15𝑋14𝑋13𝑋12! 15𝑋14𝑋13 2730
=
=
=
= 𝟒𝟓𝟓 𝒘𝒂𝒚𝒔
3! (15 − 3)!
(3𝑋2𝑋1)(12!)
3𝑋2𝑋1
6
Other problems in permutation and combination:
1. In how many ways can you arrange 3 books on a shelf from a group of 7?
n=7
r=3
7𝑃3 =
7!
7𝑋6𝑋5𝑋4!
=
= 7𝑋6𝑋5 = 𝟐𝟏𝟎 𝒘𝒂𝒚𝒔
(7 − 3)!
4!
2. In how many ways can we arrange 5 books on a shelf?
n=5
5𝑃5 = 5! = 5𝑋4𝑋3𝑋2𝑋1 = 𝟏𝟐𝟎 𝒘𝒂𝒚𝒔
3. How many teams of 4 can be produced from a pool of 12 engineers?
n = 12
r=4
12𝐶4 =
12!
12𝑋11𝑋10𝑋9𝑋8! 12𝑋11𝑋10𝑋9 11,880
=
=
=
= 𝟒𝟗𝟓 𝒘𝒂𝒚𝒔
4! (12 − 4)!
(4𝑋3𝑋2𝑋1)(8!)
4𝑋3𝑋2𝑋1
24
4. In how many ways can two same prizes be awarded among 10 contestants if both may not be given to the
same person?
n = 10
r=2
10!
10𝑋9𝑋8!
10𝑋9 90
=
=
=
= 𝟒𝟓 𝒘𝒂𝒚𝒔
2! (10 − 2)! (2𝑋1)(8!)
2
2
10𝐶2 =
5. In how many ways can two different prizes be awarded among 10 contestants if both may not be given to
the same person?
n = 10
r=2
10𝑃2 =
10!
10𝑋9𝑋8!
=
= 10𝑋9 = 𝟗𝟎 𝒘𝒂𝒚𝒔
(10 − 2)!
8!
6. A password of 6 digits is made of digits 926002. How many possible passwords are there?
𝑛1 = 1, 𝑛2 = 2, 𝑛3 = 1, 𝑛4 = 2; 𝑛 = 6
6!
= 𝟏𝟖𝟎 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒑𝒂𝒔𝒔𝒘𝒐𝒓𝒅𝒔
1! (2!)(1!)(2!)
When to Add and Multiply:
a. Whenever we come a across a situation involving 2 or more events and both events cannot occur
simultaneously, then in that case, we will simply add up all the events.
b. Whenever we come across a situation involving 2 or more events and each event can happen
simultaneously, i.e., event 1, event 2, event 3, and so on, all can happen simultaneously.
7. How many ways we can select 4 balls of same color from a basket that contains 6 black and 5 white balls?
For 4 black balls:
n=6
r=4
6𝐶4 =
6!
6𝑋5𝑋4!
6𝑋5 30
=
=
=
= 𝟏𝟓 𝒘𝒂𝒚𝒔
4! (6 − 4)! (4!)(2𝑋1) 2𝑋1
2
For 4 white balls:
n=5
r=4
5𝐶4 =
5!
5𝑋4!
5
=
= = 𝟓 𝒘𝒂𝒚𝒔
4! (5 − 4)! (4!)(1!) 1
Therefore, getting four black balls cannot occur simultaneously with getting four white balls, so we simply add
up the two events.
15 + 5 = 20 ways
8. There are 4 Czech and 3 Slovak books on the bookshelf. Czech books should be placed on the left side of
the bookshelf and Slovak books on the right side of the bookshelf. How many ways are there to arrange
the books?
Arranging Czech Books: 4P4 = 4! = 24
Arranging Slovak Books: 3P3 = 3! = 6
Therefore, arranging 4 Czech books black balls can occur simultaneously with arranging 3 Slovak books, so
we simply multiply the two events.
24X6 =144 ways
9. How many groups of 3 hearts and 5 spades can be made from the 13 hearts and 13 spades in a deck of
cards?
Choose 3 of 13 hearts:
n = 13
r=3
13C3 = 286
Choose 5 of 13 hearts:
n = 13
r=5
13C5 = 1287
Multiply the result: 286 X 1287 = 368,082 ways
10. A student must make 4 university entry exams. For passing each exam he gets either 2,3, or 4 points. He
needs to reach a least 13 points to get to the university. In how many ways can he do the exams to be
successful?
He needs to get 13, 14, 15, or 16 points.
13 points:
(3,3,3,4) 𝑛 = 4, 𝑛1 = 3, 𝑛2 = 1:
4!
=4
(3!)(1!)
(4,4,3,2) 𝑛 = 4, 𝑛1 = 2, 𝑛2 = 1, 𝑛3 = 1:
14 points:
(4,4,4,2) 𝑛 = 4, 𝑛1 = 3, 𝑛2 = 1:
(4,4,3,3) 𝑛 = 4, 𝑛1 = 2, 𝑛2 = 2:
15 points:
(4,4,4,3) 𝑛 = 4, 𝑛1 = 3, 𝑛2 = 1:
16 points:
(4,4,4,4) 𝑛 = 4, 𝑛1 = 4:
4!
= 12
(2!)(1!)(1!)
4!
=4
(3!)(1!)
4!
=6
(2!)(2!)
4!
=4
(3!)(1!)
4!
=1
(4!)
Add the results: N = 4 + 12 + 4 + 6 + 4 + 1 = 31 ways of reaching at least 13 points
11. There are 10 books kept on a bookshelf. Out of those 10, 3 are on Physics and 7 are on Chemistry. In how
many ways can Harry select 5 books from the shelf such that he selects at least one book on Physics and
one on Chemistry
Solution 1:
Physics
1
2
3
Chemistry
4
3
2
Physics
3C1 = 3
3C2 = 3
3C3 = 1
X
X
X
Chemistry
7C4 = 35
7C3 = 35
7C2 = 21
Add the
results
=
105
105
21
105+105+21
= 231 ways
Solution 2:
Select 5 books from 10 books: 10C5
The event is that you must at least select one book on physics and one on chemistry.
We have the potential to choose 5 Chemistry books, although this isn't the event, so we subtract it from
the total combinations of selecting 5 books out of 10 books, denoted as 10C5.
Select 5 Chemistry books from 7 books: 7C5
Use Subtraction:
10C5 – 7C5 = 231 ways
Probability using Permutation and Combination:
1. A Congressional committee consists of 8 men and 10 women. A subcommittee of 4 people is set up at
random. What is the probability that it will consist of all men?
The event we are interested in is all subcommittee members are males. Since there are 8 men in the committee,
the number of 4-person male subcommittees is.
Conditional Event Outcomes (Desired Outcomes): 8C4
Sample Space Outcomes (Total Possible Outcomes): 18C4
𝑃=
8𝐶4
𝟕
=
𝒐𝒓 𝟎. 𝟎𝟐𝟐𝟗 𝒐𝒓 𝟐. 𝟐𝟖%
18𝐶4 𝟑𝟎𝟔
2. Tabitha and her friend Jamie are in gym class together. Today’s class is going to involve a game of dodgeball,
where half the class is playing against the other half. If there are 30 students in the class what is the
probability that Tabitha and Jamie are on the same dodgeball team?
Selecting 15 students from 30 without arranging: 30C15
Condition: Tabitha and Jamie are on the same dodgeball team
We’ll have to consider two separate events for this because they could both be on team 1 or they could both be on
team 2.
Team 1 with Tabitha and Jamie: 28C13
Notice how the 30 has become a 28. This is because Tabitha and Jamie are already assigned to the first team,
leaving only 28 more students. Secondly, the 15 has become a 13. This is because we only need to select 13 more
students for team 1.
Team 1 without Tabitha and Jamie, meaning they’re together on Team 2: 28C15
In this case, we still needed to select 15 students for team 1, but since we can't select Tabitha or Jamie, we only
have 28 students to form our first team.
Because only one of these things will happen, we're going to add these values in our final calculation instead of
multiplying. Some math resources refer to this as an "OR" (meaning one OR the other will occur).
Conditional Event Outcomes: 28C13 + 28C15
Sample Space Outcomes: 30C15
𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
28𝐶13 + 28𝐶15 𝟏𝟒
=
𝒐𝒓 𝟎. 𝟒𝟖𝟑 𝒐𝒓 𝟒𝟖. 𝟑%
30𝐶15
𝟐𝟗
3. What is the probability of guessing a 5-digit number if none of the digits are repeated?
Conditional event outcomes: 1
Sample Space Outcomes: 10P5
𝑃=
1
𝟏
=
𝒐𝒓 𝟎. 𝟎𝟎𝟑𝟗𝟔 𝒐𝒓 𝟎. 𝟑𝟗𝟔%
10𝐶5 𝟐𝟓𝟐
4. Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.
For the numerator (conditional event outcomes), we need the number of ways to draw one Ace and four other
cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will
be 4C1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48C4 ways to
select the four non-Aces.
Conditional Event Outcomes: 4C1 X 48C4
Sample Space Outcomes: 52C5
𝑃=
(4𝐶1)(48𝐶4)
𝟕𝟕𝟖, 𝟑𝟐𝟎
=
𝒐𝒓 𝟎. 𝟐𝟗𝟗𝒐𝒓 𝟐𝟗. 𝟗%
52𝐶5
𝟐, 𝟓𝟗𝟖, 𝟗𝟔𝟎
5. In a certain state’s lotter, 48 balls numbered 1 through 48 are placed in machine and six of them are drawn
at random. If the six numbers drawn match the numbers that a player had chosen, the player wins Php
10,000,000. In this lotter the, the order the numbers are drawn in doesn’t matter. Compute the probability
that you win the million-dollar prize if you purchase a single lottery ticket.
The number of ways to choose 6 out of the 6 winning numbers is given by 6C6 = 1
Conditional Event Outcomes: 6C6
Sample Space Outcomes (Total Possible Outcomes): 48C6
𝑃=
6𝐶6
𝟏
=
𝒐𝒓 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟖𝟏𝟓
48𝐶6 𝟏𝟐, 𝟐𝟕𝟏, 𝟓𝟏𝟐
6. Every student at Shellie’s school is assigned a 4-digit PIN to access the school’s computers. The first digit
of the code is 0 or 1. The remaining digits are chosen from the numbers 2 through 9, and no number may
be used more than once in a code. What is the probability that Shellie is assigned the PIN 1234?
There is only 1 favorable arrangement, 1234.
Conditional Event Outcomes: 1
For the first digit, there are 2 numbers to pick from 0 or 1: 2P1 = 2
For the last three digits, there are 8 numbers to pick from, 2 through 9: 8P3 = 336
Multiply: 2X336 = 672 possible pins
Sample Space Outcomes: 672
𝑷=
𝟏
𝒐𝒓 𝟎. 𝟎𝟎𝟏𝟒𝟗 𝒐𝒓 𝟎. 𝟏𝟒𝟗%
𝟔𝟕𝟐
7. A shipment of 20 cars contains 3 defective cars. Find the probability that a sample size of 2, drawn from
the 20, will not contain exactly one defective car.
For the sample to contain exactly one defective car, a car must be selected from the defective cars and another
from the non-defective cars.
One Defective car: 3C1
Non-Defective cars: 17C1
Sample Space Outcomes - Multiply: 3C1 X 17C1
Start by determining how many ways there are to select a sample of 2 cars from 20 cars.
Sample Space Outcomes: 20C2
𝑃=
(3𝐶1)(17𝐶1)
𝟓𝟏
=
𝒐𝒓 𝟎. 𝟐𝟔𝟖 𝒐𝒓 𝟐𝟔. 𝟖%
20𝐶2
𝟏𝟗𝟎
8. Suppose three people are in a room. What is the probability that there is at least one shared birthday
among these three people?
What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no
shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is
a complementary event,
P(shared birthday) = 1 – P(no shared birthday)
We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are
one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of
365 for your birthday. What is the probability that the second person does not share your birthday? There
are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364
choices that will guarantee that you do not share a birthday with this person, so the probability that the
second person does not share your birthday is 364/365. Now we move to the third person. What is the
probability that this third person does not have the same birthday as either you or the second person?
There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the
third person does not share a birthday with the first two is 363/365.
365
364
363
P(no shared birthday) = (365) (365) (365) = 0.9918
P(shared birthday) = 1 – 0.9918 = 0.0082 or 0.82%