Chapter 15
Diagrams, Principals, Permutations,
Combinations and Binomial Theorem
Mr. Morrow
4/9/2013 – 4/19/2013
- Chapter Objectives To use Venn diagrams to illustrate intersections and unions of sets and to
use the inclusion-exclusion principle to solve counting problems involving
intersections and unions of sets.
To use the multiplication, addition and complement principles to solve
counting problems.
To solve problems involving permutation and combinations
To solve counting problems that involve permutations with repetition and
circular permutations
To use the binomial theorem and Pascal’s triangle
- Combinatorics The theory of counting:
- Counting should be easy but it can get tricky when there are
numerous objects to be counted
- Shortcuts (rules) make it easier to count large numbers of objects
- Venn Diagrams These diagrams, in which you all should be familiar with, can be used to
illustrate and separate sets of number we are counting
Definitions:
- The Universal Set is the set of ‘objects’ or elements to be
counted and is often represented by the letter U
The number of elements in the universal set is
represented by n(U)
In a Venn Diagram the universal set is
represented as a rectangle
- A subset is the group of elements that we are interested in
In a Venn Diagram each subset is represented by a circle
- Class Poll -
Who has a brother or brothers?
Who has a sister or sisters?
Who has both brothers and sisters?
Who has neither brothers nor sisters?
We can represent this information using
a Venn Diagram
Who wishes they didn’t have their brothers or
sisters??
- Venn Diagram -
U
B
n(B) =
n(U) =
S
n(S) =
- Venn Diagram The intersection of the subsets are
the people with brothers and sisters.
U
B
S
n(S) =
n(B) =
n(B  S) =
- Venn Diagram The union of the subsets represents
the people are in either set
U
S
B
Union -
B S
- The Inclusion – Exclusion Principle U
B
S
Because the intersection is counted in each subset, when
we determine the union we only count the intersection
once therefore:
𝒏 𝑩 ∪ 𝑺 = 𝒏 𝑩 + 𝒏 𝑺 − 𝒏(𝑩 ∩ 𝑺)
- The Complement Principle U
B
S
The compliment of set B is written as either B’ or 𝐵 and
is the elements not in set B:
𝒏 𝑩 ′ = 𝒏 𝑼 − 𝒏(𝑩)
- Complement of the Union -
The Confederacy! Haha
- Complement of the Union The complement of the subsets is the number of people
who have neither brothers nor sisters
U
n(B  S)
B
S
n(S) =
n(B) =
n(U) =
n(B  S) =
𝑛(𝐵 ∪ 𝑆) = 𝑛 𝑈 − [𝑛 𝐵 + 𝑛 𝑆 − 𝑛 𝐵 ∩ 𝑆 ]
- Lets wrap it up -
U
n(B  S) 
B
S
n(S) =
n(B) =
n(U) =
n(B  S) =
- Example Of the 540 seniors at Central High School, 335 are
taking math, 287 are taking science and 220 are taking
both math and science. How many are taking neither math
or science?
- Example U
M
n(M) = 335
S
n(S) = 287
n(U) = 540
n(M  S) = 220
The number of students taking either math or science can be determined
using the inclusion – exclusion principle.
n(M  S)  n(M)  n(S)  n(M  S)
= 335 + 287 – 220
= 402
To find the number of seniors taking neither math or science we find the
complement of the union which is n(M  S) = 540 – 402 = 138
- Practice Given the set of number from 1 to 30. Determine the
subsets of even numbers and numbers divisible by 3.
Represent this information in a Venn Diagram.
- Determine 𝑛(𝐸 ∩ 𝐷)
- Determine 𝑛(𝐸 ∪ 𝐷)
- Determine 𝑛(𝐸)′
- Determine 𝑛(𝐸 ∪ 𝐷)′
- Practice -
25
5
1
11
16
2
7
8
14
4
26
6
22
12
20
18
24
30
10
3
27
9
19
15
21
28
23
17
13
29
- Practice -
𝑛 𝐸 ∪ 𝐷 = 20
𝑛 𝐸∩𝐷 =5
- Practice 𝑛 𝐸 ′ = 15
- Practice 𝑛 𝐸 ∪ 𝐷 ′ = 10
- Practice If A={3, 6, 9, 12, 15, 18}, B={2, 4, 6, 8, 10, 12, 14, 16, 18} and
C={1, 4, 9, 16} list the elements in
A  B {6, 12, 18} B  C {4,16}
A  C {9}
A  B  C {} A  B
A  C {1, 3, 4, 6, 9, 12, 15, 16, 18}
{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18}
BC
ABC
(A  B)  C {4, 9, 16}
{1, 2, 4, 6, 8, 9, 10, 12, 14, 16, 18}
(B  C)  A (A  C)  B
{3, 4, 6, 9, 12, 15, 16, 18}
{2, 4, 6, 8, 9, 10, 12, 14, 16, 18}
Homework: pg. 568 (# 1, 3, 11, 13, 15, 17, 19, 21)
- Homework Questions -
- Multiplication Principle The other day we saw how we can use Venn Diagrams to help us
visualize concepts of counting. Right now we will see an idea
called “Trees”
Example:
Try flipping a coin three times, how many possible outcomes
are there?
H
H
H
T H
T
H
T
T H
T
H
T
There are eight possible outcomes:
HHH, HHT, HTH, HTT,
THH, THT, TTH, TTT
- Multiplication Principle Definition: If an action can be performed in n1 ways and for each of
these ways another action can be performed in n2 ways, then the
two actions can be performed in n1n2 ways.
Practice:
How many of you girls stand in your
closet for at least 5 minutes a day?
This is widely not true, but assume
you only have three button-ups,
two blazers and five pairs of pants.
How many outfits are available?
Draw a tree to verify
- Multiplication Principle It is possible to select from the 3 shirts, 2 blazers and 5 pants by
applying the multiplication principle:
Outfits = 3 ∙ 2 ∙ 5 = 30
- Practice The license plate math: In Virginia a standard (none custom)
license plate is issued like the one below. How many
combinations can be created?
Think about what you see before you answer…
___ ∙ ___ ∙ ___ ∙ ___ ∙ ___ ∙ ___ ∙ ___
26 ∙ 26 ∙ 26 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 175,760,000
- Addition Principle We just saw how many possible license plates could be made
using the picture below. Now consider that VA may decide to
start making licenses such that the 1st 3 spaces are numbers
and the last 4 spaces are letters. How many possible licenses
would VA now have?
(3 letter/ 4 numbers)
26 ∙ 26 ∙ 26 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 175,760,000
(3 numbers/ 4 letters)
10 ∙ 10 ∙ 10 ∙ 26 ∙ 26 ∙ 26 ∙ 26 = 456,976,000
Ok…now what?
This example introduces the concept of mutually exclusive
events. We can’t put both a letter and a number into each space
– mutually exclusive events cannot be performed together.
- Mutually Exclusive Venn Diagrams How do we represent mutually exclusive events?
U
L
N
The addition principle states that if two actions are mutually
exclusive, and the first can be done n1 ways and the second can
be done n2 ways, the one action or the other can be done in
n1 + n2 ways.
- Addition Principle Soo…how many possible licenses would VA now have?
(3 letter/ 4 numbers)
26 ∙ 26 ∙ 26 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 175,760,000
(3 numbers/ 4 letters)
10 ∙ 10 ∙ 10 ∙ 26 ∙ 26 ∙ 26 ∙ 26 = 456,976,000
175,760,000 + 457,976,000 = 632,736,000
- Factorials How many ways can I arrange the Harry Potter series books on a
shelf?
… The first thing we need to know is how many books
are in the series?
Ok, now think of having 7 slots on the shelf
7
6
5
How many books could go in
the first slot? second? third?
Why?
We multiply each by the
preceding term…
7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 7!
= 5040
4
3
2
1
- Practice 1.
If 8! is 40,320 what is 9! How would we determine 23!
• Go to the <run> menu on your calculator. Press the
OPTN button. Press <F6> which is the arrow and then
press <F3> which is PROB. Enter the factorial number
then press <F1> to invoke the factorial calculation
9! = 362,880 and 23! = 2.58 x 1022
2. If Costa Rica chose not to use 0 on its license plates, how
many different plates are possible?
Costa Rican tags = 9 x 9 x 9 x 9 x 9 = 59049
3. A guy has six ball caps and 3 hoodies
1. How many cap and hoodie combinations can he choose
from?
2. If he chooses to wear a ball cap or a hoodie but not both,
how many choices does he have?
Outfits = 6 x 3 = 18 or 6 + 3 = 9
4. If 10 runners compete in a race, in how many ways can 1st,
2nd and 3rd place prizes be awarded?
Prizes = 10 x 9 x 8 = 720
- Homework Page 575 – 577 (# 1, 3, 5, 7, 11, 13, 15, 17)
- Warm Up In an election-day survey of 100 voters leaving the polls, 52
said they voted for Proposition 1, and 38 said they voted for
Proposition 2. If 18 said they voted for both, how many
voted for neither?
In a survey of 48 high school students, 20 liked classical
music and 16 liked bluegrass music. Twenty students said
they didn’t like either. How many liked classical but not
bluegrass?
For a universal set U, what is 𝑈?
- Homework Questions -
- Permutations Given n symbols, how many non-repetitive lists of length r can be made
from the n symbols in which order matters? By using the multiplication
principle to obtain the answer
𝑛 𝑛 − 1 𝑛 − 2 … (𝑛 − 𝑟 + 1)
By cancellation this value can also be written as:
𝑛 𝑛 − 1 𝑛 − 2 … (𝑛 − 𝑟 + 1)(𝑛 − 𝑟)(𝑛 − 𝑟 − 1) … 3 ∙ 2 ∙ 1
𝑛 − 𝑟 ∙ 𝑛 − 𝑟 − 1 …3 ∙ 2 ∙ 1
𝑛!
=
𝑛−𝑟 !
We summarize this as follows:
The number of non-repetitive lists of length r whose entries
𝑛!
are chosen from a set of n possible entries is
.
𝑛−𝑘 !
- Permutations For this class, you will represent this as:
𝑛!
𝑛𝑃𝑟 =
𝑛−𝑟 !
Where:
n is the number of objects/ elements available to be chosen
r is the number of objects/ elements actually chosen
- Combinations Without diving too deep into the math behind this next idea…
Imagine we build from Permutations but instead we are curious about
solutions in which order does not matter.
For this class, you will represent this as:
𝑛!
𝑛𝐶𝑟 =
𝑛 − 𝑟 ! 𝑟!
Where:
n is the number of objects/ elements available to be chosen
r is the number of objects/ elements actually chosen
- Examples A company advertises two job openings, one for a computer
programmer and one for an IT specialist. If 10 people who are
qualified for either position apply, in how many ways can the job
openings be filled?
A company advertises two job openings
for computer programmers, both with
the same salary and job description.
In how many ways can the openings be
filled if 10 people apply?
- Examples A company advertises two job openings, one for a computer
programmer and one for an IT specialist. If 10 people who are
qualified for either position apply, in how many ways can the job
openings be filled?
10!
10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
=
10𝑃2 =
10 − 2 !
8∙7∙6∙5∙4∙3∙2∙1
= 90
A company advertises two job openings for computer programmers,
both with the same salary and job description. In how many ways can
the openings be filled if 10 people apply?
10!
10𝐶2 =
10 − 2 ! 2!
= 45
- Using a Calculator Your calculator will quickly compute permutations and combinations.
For example 1 we are attempting to calculate 10P2
Go to the Run menu
Press the OPTN button
Press F6 (arrow over to additional menu choices)
Press F3 (PROB)
Key in 10, press F2 (nPr) and key in 2
Press EXE
For example 2 we are attempting to calculate 10C2
Go to the Run menu
Press the OPTN button
Press F6 (arrow over to additional menu choices)
Press F3 (PROB)
Key in 10, press F3 (nCr) and key in 2
Press EXE
- Practice 1) A single 5-card hand is dealt off of a standard 52-card deck. How
many different 5-card hands are possible?
2) Same scenario are (1), however, how many such hands are there in
which two of the cards of clubs and three are hearts?
3) Imagine a lottery that works as follows. A bucket contains 36 balls
numbered 1, 2, 3, 4, …, 36. Six of these balls will be drawn
randomly. For $1 you buy a ticket that has six blanks. You fill in the
blanks with six different numbers between 1 and 36. You win
$1,000,000 if you chose the same numbers that are drawn,
regardless of the order. What are the changes of winning?
- Practice A single 5-card hand is dealt off of a standard 52-card deck. How many
different 5-card hands are possible?
52!
52𝐶5 =
52 − 5 ! 5!
= 2,598,960
Same scenario are (1), however, how many such hands are there in
which two of the cards of clubs and three are hearts?
Think of such hand as being described by a list of length two of the
form where the first entry is a 2-element subset of 13 club cards, and
the second entry is a 3-element subset of the 13 heart cards.
13!
13!
( 13𝐶2 )( 13𝐶3 ) =
∙
13 − 2 ! 2! 13 − 3 ! 3!
= 22,308
- Practice Imagine a lottery that works as follows. A bucket contains 36 calls
numbered 1, 2, 3, 4, …, 36. Six of these balls will be drawn randomly.
For $1 you buy a ticket that has six blanks. You fill in the blanks with six
different numbers between 1 and 36. You win $1,000,000 if you chose
the same numbers that are drawn, regardless of the order. What are
the changes of winning?
You are choosing six numbers from a set of 36 numbers
36!
36𝐶6 =
36 − 6 ! 6!
= 1,947,792
Homework: pg. 580 (# 1-19 odd)
- Ummm?? Write down all the possible arrangements of the letters MOP
How many arrangements were you able to create?
Write down all the distinguishable arrangements of the letters MOM
How many arrangements were you able to create?
Why was the total number of arrangements
different for the two sets?
- Permutations with Repetition If an element is repeated in an arrangement, then fewer permutations
result.
𝑛!
P=
𝑛1 ! 𝑛2 ! 𝑛3 ! … 𝑛𝑘 !
Where:
n1 = the number of elements of type 1,
n2 = the number of elements of type 2 …
nk = the number of elements of type k.
This gives us the number of distinguishable permutations of n elements
- Example Consider MEXICO and CANADA
Each of the letters in Mexico is different
6𝑃6
= 6! = 720
A is repeated three times in CANADA
P=
6𝑃6
3!∙1!∙1!∙1!
= 120
- Example How many permutations are there for the letters of
MASSACHUSETTS
13!
= 64,864,800
4! 2! 2! 1! 1! 1! 1!
- Practice A person is at point X on the grid below and is going to walk to
point Y by always traveling south or east. How many routes from
X to Y are possible?
X
Y
10!
= 210
4! 6!
- Wait a minute? How many circular permutations are possible when seating four
people around a table?
B
A
A
C
D
D
B
C
C
A
B
D
D
C
B
A
ABCD
DABC
CDAB
BCDA
• Circular permutations are the same because A
is always to the right of B which is to the right of C which
is to the right of D.
• Linear permutations are different
Therefore the number of circular permutations =
𝑛!
𝑛
- Practice 1. How many different ways can five children arrange
themselves for a game of ring-around-the-rosie?
2. How many ways can ten people
be seated around a circular table
if the host and hostess can’t be
seated together?
Homework pg. 585 (#1-15 odd)
- Practice -
1. P = 5!
5
P = 24
10
7
2.
8
7
1
2
6
3
5
4
10 ×8 × 7 × 7!
P=
10
P = 282,240
- Warm Up -
- Homework Questions -
- Investigation Determine:
𝑎+𝑏
(𝑎 + 𝑏)1 =
𝑎2 + 2𝑎𝑏 + 𝑏 2
(𝑎 + 𝑏)2 =
𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3
(𝑎 + 𝑏)3 =
𝑎2 + 4𝑎3 𝑏 + 6𝑎2 𝑏 2 + 4𝑎𝑏 3 + 𝑏 4
(𝑎 + 𝑏)4 =
(𝑎 + 𝑏)5 = 𝑎5 + 5𝑎4 𝑏 + 10𝑎3 𝑏 2 + 10𝑎2 𝑏 3 + 5𝑎𝑏 4 + 𝑏 5
Can you see any patterns?
How do the number of terms relate to the exponent?
Is there a pattern to the exponents in each term?
Do you detect any relationship for the coefficients of each
term?
- Aint nobody got time to do all that Our objective now is to develop a method for expanding
binomials of the (𝑎 + 𝑏)𝑛 where n is a positive integer
When we expand (FOIL) the linear factors in a binomial
the result is a polynomial with n+1 terms
Each term in the polynomial is of the form 𝑘𝑎 𝑦 𝑏 𝑧 where k
is a coefficient and y and z are exponents.
The hard part is finding a simple way to determine k, y, and z.
- Binomial Theorem If n is a positive integer then:
(𝑎 + 𝑏)𝑛 = 𝑛𝐶0 𝑎𝑛 𝑏 0 + 𝑛𝐶1 𝑎𝑛−1 𝑏1 + 𝑛𝐶2 𝑎𝑛−2 𝑏2 + ⋯ + 𝑛𝐶𝑛 𝑎0 𝑏 𝑛
Where each term is in the form:
𝑘𝑎 𝑦 𝑏 𝑧
- Therefore for any term t of a binomial to the 𝑛𝑡ℎ power the
coefficient 𝑘 = 𝑛𝐶𝑡−1
- The sum of the exponents of a and b is always n. (y + z = n)
- The exponent of the first monomial of ‘a’, y = n – (t – 1)
- The exponent of the second monomial of ‘b’, z = t – 1
- Example Find the first four terms in the expansion of (𝑎 + 𝑏)21
1st term
k = 21C(1-1) = 21C0 = 1
y = 21 – (1 – 1) = 21
z=1–1=0
therefore the first term is the product of 1 x a21 x b0 = a21
2nd term
k = 21C(2-1) = 21C1 = 21
y = 21 – (2 – 1) = 20
z=2–1=1
therefore the second term is 21a20b1 = 21a20b
3rd term
k = 21C(3 – 1) = 21C2 = 210
y = 21 – (3 – 1) = 19
z=3–1=2
therefore the third term is 210a19b2
4th term
k = 21C(4 – 1) = 21C3 = 1330
y = 21 – (4 – 1) = 18
z=4–1=3
therefore the fourth term is 1330a18b3
- Example What is the 8th term of (𝑥 + 2𝑦)10 ?
1. k = 10C8-1
k = 10C7
k = 120
2. The exponent for x = 10 – (8 – 1) = 3
3. The exponent for -2y = 8 – 1 = 7
4. Therefore the 8th term is
= 120x3(-2y)7
= 120x3(-2)7y7
= 120(-128)x3y7
= - 15360x3y7
- Practice 1.
Give the expansion of (2a – b2)8
2.
In the expansion of (a + b)20, what is the coefficient of
a) a17b3
b) a3b17
3.
Find the first four terms of the expansion of the expression
(sinx – cosy)30. Do not simplify.
4.
Find the value of (0.99)5 to the nearest hundredth by
considering the expansion of (1 – 0.01)5
- Practice Expanding (2a – b2)8 we know that there are 8 + 1 = 9 terms.
1st term = 8C0(2a)8(-b2)0 = 256a8
2nd term = 8C1(2a)7(-b2)1 = (8)(-128)a7b2 = -1024a7b2
3rd term = 8C2(2a)6(-b2)2 = (28)(64)a6b4 = 1792a6b4
4th term = 8C3(2a)5(-b2)3 = (56)(-32)a5b6 = -1792a5b6
5th term = 8C4(2a)4(-b2)4 = (70)(16)a4b8 = 1120a4b8
6th term = 8C5(2a)3(-b2)5 = (56)(-8)a3b10 = -448a3b10
7th term = 8C6(2a)2(-b2)6 = (28)(4)a2b12 = 112a2b12
8th term = 8C7(2a)1(-b2)7 = (8)(-2)ab14 = -16ab14
9th term = 8C8(2a)0(-b2)8 = (1)(1)b16 = b16
(2a – b2)8 = 256a8 - 1024a7b2 +1792a6b4 -1792a5b6 +1120a4b8
-448a3b10 +112a2b12 -16ab14 +b16
- Practice In the expansion of (a + b)20, what is the coefficient of
a) a17b3
b) a3b17
a) y = n – (t – 1)
17 = 20 – (t – 1)
t=4
k = nCt-1
k = 20C3
k = 1140
b) y = n – (t – 1)
3 = 20 – (t – 1)
t = 18
k = nCt-1
k = 20C17
k = 1140
- Practice Find the first four terms of the expansion of the expression
(sinx – cosy)30. Do not simplify.
1st term = 30C0(sinx)30(-cosx)0 2nd term = 30C1(sinx)29(-cosx)1
3rd term = 30C2(sinx)28(-cosx)2 4th term = 30C3(sinx)27(-cosx)3
1st Term
k = 30C0 = 1
a = 30 – (1 – 1) = 30
b=1–1=0
kxaby = sin30x
2nd Term
k = 30C1 = 30
a = 30 – (2 – 1) = 29
b=2–1=1
kxaby = -30sin29xcosy
3rd Term
k = 30C2 = 435
a = 30 – (3 – 1) = 28
b=3–1=2
kxaby = 435sin28xcos2y
4th Term
k = 30C3 = 4060
a = 30 – (4 – 1) = 27
b=4–1=3
kxaby = -4060sin27xcos3y
- Practice Find the value of (0.99)5 to the nearest hundredth by considering
the expansion of (1 – 0.01)5
(1 – 0.01)5 = 5C0(1)5(-.01)0 + 5C1(1)4(-.01)1 + 5C2(1)3(-.01)2 +
2
3
1
4
0
5
5C3(1) (-.01) + 5C4(1) (-.01) + 5C5(1) (-.01)
= 1 - .05 + .001 – .00001 + .00000005 - .0000000001
= 0.9509900499
Homework: pg. 592 (#1-21 odd)