Download Permutations and Combinations Student Notes

PERMUTATIONS AND COMBINATIONS
1. Fundamental Counting Principle
Assignment: Workbook: pg. 375 – 378 #1-14
2. Permutations and Factorial Notation
Assignment: Workbook pg. 382-384 #1-13, pg. 526 of text #22
3. Permutations with Repetitions and Restrictions
Assignment: Workbook: pg. 388 – 391 #1-15, pg. 525 of text #8
4. Combinations – pg. 528-536
Assignment: Workbook: pg. 396 –398 #1-12, pg. 402-404 #1-16, pg. 534 of text #6
5. The Binomial Theorem
Assignment: Workbook: pg. 410 – 412 #4, 7-13, pg. 415 #1abcdef
6. Chapter Quiz
7. Chapter Review – pg. 546-547
Assignment: pg. 546-547 #1-12, 14-18
8. Chapter Exam
FUNDAMENTAL COUNTING PRINCIPLE
Learning Outcomes:

To solve problems using the fundamental counting principle

To determine, using a variety of strategies, the number of permutations of n elements
taken r at a time.
Ex. A toy manufacturer makes a wooden toy in three parts:
Part 1: the top part may be coloured red, white or blue
Part 2: the middle part may be orange or black,
Part 3: the bottom part may be yellow, green, pink, or purple
How many different parts are there to the toy?
For each part of the toy, how many choices of colours do you have?
Multiply all the choices together, now you have the total amount of possible toys:
____
top
.
___
.
middle
____
bottom
= ____
coloured toys
Check your answer using a tree diagram:
Make a conjecture about how you can use multiplication only to arrive at the number of different
coloured toys possible.
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The Fundamental Counting Principle
Consider a task made up of several stages. If the number of choices for the first stage is a, the
number of choices for the second stage is b, the number of choices for the third stage is c, etc.,
then the number of ways in which a task can be completed is a x b x c x ….. This is called the
fundamental counting principle.
Ex. Determine the number of distinguishable four letter arrangements that
can be formed from the word ENGLISH if:
a. letters can be repeated?
b. no letters are repeated and:
i)
there are no further restrictions?
ii)
the first letter must be E?
iii)
the “word” must contain G?
iv)
the first and last letters must be vowels?
Ex. The telephone numbers allocated to subscribers in a rural area consist
of one of the following:
- the digits 345 followed by any three further digits or
- the digit 2 followed by one of the digits 1 to 5, followed by any three further
digits.
How many different telephone numbers are possible?
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Ex. Car number plates in an African country consist of a letter other than I
or O followed by three digits, the first of which cannot be zero,
followed by any two letters which are not repeated. How many
different car number plates can be produced?
Ex. Consider the digits 2, 3, 5, 6, 7 and 9.
a) If repetitions are not permitted, how many 3-digit number can be
formed?
b) How many of these are:
i) less than 400?
ii) even?
iii) multiples of 5?
Workbook: pg. 375 – 378 #1-14
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PERMUTATIONS AND FACTORIAL NOTATION
Learning Outcomes:

To solve problems using the fundamental counting principle

To determine, using a variety of strategies, the number of permutations of n elements
taken r at a time.

To solve an equation that involves n Pr notation
Factorial Notation
Consider how many ways there are of arranging 6 different books side by side on a shelf. In this
example we have to calculate the product 6x5x4x3x2x1. In mathematics this product is denoted
by 6! (“factorial” or “factorial 6”) In general n!=n(n-1)(n-2)(n-3)….(3)(2)(1), where n W
Ex. Use the factorial key on your calculator to solve 6!
10!
, there are several approaches:
7!
a.Use you calculator:
Ex. To simplify
b. By Cancellation:
10!
7!
43!
Ex. Find the value of 40!
Ex. Simplify the following expressions:
a.
b.
𝑛!
(𝑛−2)!
(𝑛+3)!
𝑛!
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Permutations
An arrangement of a set of objects in which the order of the objects is important is called a
permutation.
Ex. How many permutations are there of the letters of the word:
a. REGINA:
b. KELOWNA:
The number of permutations of “n” different objects taken “r” at a time is:
n!
n Pr  (n  r )!
Ex. Use the n Pr key on your calculator to evaluate 8 P3 . Then solve using
factorials.
8 : Math  PRB  2  3  336
Using Factorials:
n
pr 
n!
8!
8! 8  7  6  5!

 
 8  7  6  336
(n  r )! (8  3)! 5!
5!
Defining 0!
If we replace r by n in the above formula we get the number of permutations of n objects taken n
at a time. This we know is n!
n
pn  n ! 
n!
n!

(n  n)! 0!
For this to be equal to n! the value of 0! Must be 1.
0! Is defined to have a value of 1
Try it:
Solve 2𝑃2
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Ex. In a South American country, vehicle license plates consist of any 2
different letters followed by 4 different digits. Find how many different
license plates are possible using:
fundamental counting principle:
permutations:
(letters)(digits)
Ex. Solve for n in the equation n P4  28  n 1 P2
n!
28(n  1)!

(n  4)!  (n  1)  2!
28  n  1!
n!

(n  4)!
(n  3)!
In many cases involving simple permutations, the fundamental counting principle can be used in
place of the permutation formulas.
Assignment: pg. 382-384 #1-13
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PERMUTATIONS WITH RESTRICTIONS AND REPETITIONS
Learning Outcomes:

To solve problems using the fundamental counting principle

To determine, using a variety of strategies, the number of permutations of n elements
taken r at a time.

To solve an equation that involves n Pr notation

To solve counting problems when two or more elements are identical
Permutations with Restrictions
In many problems restrictions are placed on the order in which objects are arranged. In this type
of situation deal with the restrictions first.
Ex. In how many ways can all of the letters of the word ORANGES be
arranged if:
a.)there are no further restrictions:
b.)the first letter must be an N? :
c.)the vowels must be together in the order O, A, and E:
You need to multiply by 5 because the OAE can fill into any of the 5 slots
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Ex. Find the number of permutations of the letters in the word KITCHEN if:
a.)the letters K, C, and N must be together but not necessarily in that
order
b.)the vowels must not be together
easier to find complementary event: vowels kept together, then subtract from no restrictions (7!)
7! - 2 . 1 .
5 . 4 . 3 . 2 . 1
vowels
=7! – ((2!)(5!)(6))
= 5040 – 1440 = 3600
Ex. In how many different ways can 3 girls and 4 boys be arranged in a row
if no two people of the same gender can sit together
Permutations with Repetitions
The following formula gives the number of permutations when there are repetitions:
The number of permutations of n objects, where a are the same
of one type, b are the same of another type, and c are the same
of yet another type, can be represented by the expression below
n!
a !b !c !
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Ex. Find the number of permutations of the letters of the word:
a.) VANCOUVER:
b.) MATHEMATICAL:
Ex. How many arrangements of the word POPPIES can be made under
each of the following conditions?
a.) without restrictions:
b.) if each arrangement begins with a P:
P OPPIES
The first letter is a P, and the next 6 letters can be in any arrangement (remember there is a
repetition of 2P’s within the remaining 6 letters)
c.) if all the P’s are together:
PPP OIES
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d.) if the first letter is P and the next one is not P:
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Ex. Brett bought a carton containing 10 mini boxes of cereal. There are 3
boxes of Corn Flakes, 2 boxes of Rice Krispies, 1box of Coco Pops, 1
box of Shreddies, and the remainder are Raisin Bran. Over a ten day
period Brett plans to eat the contents of one box of cereal each
morning.
How many different order are possible if on the first day he has Raisin Bran?
CF, 2 RK, 1 CP, 1 S, 3 RB (will only have 2 after first day)
Consider the following problem:
Ex. A city centre has a rectangular road system with 5 streets running
north to south and 6 avenues running west to east.
a.Draw a grid to represent this situation
Sean
b. Sean is driving a car and is situated at the extreme northwest corner of the city
centre. In how many ways can he drive to the extreme southeast corner if at
each turn he moves closer to his destination (assume all streets and avenues
allow two –way traffic)
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Ex. Find the number of pathways from A to B if paths must always move
closer to B.
A
One possible path is shown, any
path from A to B must travel 3
right, 2 down and 1 forward
B
b
“Find the number of pathways from A to B if paths must always move
closer to B”
A
B
Homework: Workbook: pg. 388 – 391 #1-15
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LESSON 4: COMBINATIONS
Learning Outcomes:

To explain the differences between a permutation and a combination

To determine the number of ways to select r elements from n different elements

To solve problems using the number of combinations of n different elements taken r at a
time

To solve an equation that involves n Cr notation
When is order not important?
1. From a group of four students, three are to be elected to an executive committee with a
specific position. The positions are as follows:
1st position
President
2nd position
Vice President
3rd position
Treasurer
a. Does the order in which the students are elected matter? Why?
b. In how many ways can the positions be filled from this group?
2. Now suppose that the just elected three students are to be selected to serve on a committee.
a. If all the committee positions are the same, does it matter if you are selected first or
third?
b. Is the order in which the four students are selected still important? Why or why not?
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c. How many committees from the original group of four students are now possible?
4
1 (President, VP, Treasurer), 2(President, VP, student 4), 3( VP, Treasurer, student 4),
4( President, Treasure, student 4)
3. You are part of a group of 6 students who are about to shake hands with every other person.
a. How many students shake hands at one time?
b. Does the order of the handshakes matter?
c. How many handshakes are possible if each student shakes every other student’s hand
once?
4. Part 1 deals with permutations, part 2 and 3 dealt with combinations. What is the biggest
difference between a permutation and a combination?
Permutations require that the order of the grouped objects is important. Combinations involve
arrangements where order of the grouped objects is not important.
A combination is a selection of a group of objects, taken from a larger group for which the kinds
of objects selected is important, but not the order in which they are selected.
There are several ways to find the number of possible combination. One is to use reasoning.
Use the fundamental counting principle and divide by the number of ways that the object can be
arranged among themselves.
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For example, calculate the number of combinations of three digits made from the digits 1, 2, 3, 4,
and 5:
5 x 4 x 3 = 60
However, 3 digits can be arranged 3! ways among themselves. So:
60
= 10
3!
Formula:
The number of combinations of “n” items take “r” at a time is:
n!
n Cr   n  r !r !
Use the formula to solve the last problem:
5!
= 10
(5 − 3)! 3!
The n Cr key on the calculator can be used to evaluate combinations:
Math  PRB  3  n Cr
 n
In some texts n Cr is written as  
r 
Ex. Three students from a class of 10 are to be chosen to go on a school trip.
In how many ways can they be selected?
Ex. To win the LOTTO 649 a person must correctly choose six numbers
from 1 to 49. Jasper, wanting to play LOTTO 649, began to wonder
how many numbers he could make up. How many choices would
Jasper have to make to ensure he had the six winning numbers?
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Ex. The Athletic Council decides to form a sub-committee of seven council
members to look at how funds raised should be spent on sports activities
in the school. There are a total of 15 athletic council members, 9 males
and 6 females. The sub-committee must consist of exactly 3 females.
a. In how many ways can the females be chosen?
b. In how many ways can the males be chosen?
c. In how many ways can the sub-committee be chosen?
Exactly 3 females
(females)(Males)
d. In how many ways can the sub-committee be chosen if Bruce the
football coach must be included?
(females)(males)(Bruce)
Ex. Consider a standard deck of 52 cards. How many different five card
hands can be formed containing:
a. at least 1 red card
find complementary even ( no restrictions – no red)
52
C5  26 C5
= 2 533 180 different five card hands
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b. at most 2 kings
3 events – 0 kings, 1 king, 2 kings
c. exactly two pairs?
1st pair x 2nd pair x 1 other card x what number for the pair
Combinations which are equivalent
Ex. Jane calculated
10
C2 to be 45 arrangements. She then calculated 10 C8 to
be 45 arrangements. Use factorial notation to prove this:
10
C2  10 C8
10!
2!8!
10!
8!2!
10  9  8!
2!8!
10  9  8!
8!2!
45
=
45
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Solving for “n” in Combinations Problems
Ex. During a Pee Wee hockey tryout, all the players met on the ice after the
last practice and shook hands with each other. How many players
attended the tryouts if there were 300 handshakes in all?
A handshake requires 2 people
n
C2  300
n!
 300
(n  2)!2!
n!
 600
(n  2)!
Polygons and Diagonals
The number of diagonals in a regular n-sided polygon is:
n C2  n
Ex. How many diagonals are there in a regular octagon?
Ex. A polygon has 65 diagonals. How many sides does it have?
n
C2  n  65
n!
 65  n
(n  2)!2!
n(n  1)(n  2)!
 2(65  n)
(n  2)!
Workbook: pg. 396 –398 #1-12, pg. 402-404 #1-1
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LESSON 5: THE BINOMIAL THEOREM
Learning Outcomes:

To relate the coefficients in the expansion of (𝑥 + 𝑦)𝑛 , n ε N, to Pascal’s triangle and to
combinations

To expand (𝑥 + 𝑦)𝑛 , n ε N, in a variety of ways, including the binomial theorem

To determine a specific term in the expansion of (𝑥 + 𝑦)𝑛
Pascal’s Triangle:
1
1
1
1
2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Notice the pattern?
Can you continue the pattern for the next two rows?
Recall Pathway questions, can you solve the following question using pascal’s triangle?
“Find the number of pathways from A to B if paths must always move
closer to B”
A
B
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How about this one?
Ex. A city centre has a rectangular road system with 5 streets running
north to south and 6 avenues running west to east.
a.Draw a grid to represent this situation
Sean
Ex. Complete the following expansions:
a. (𝑥 + 𝑦)2
b. (𝑥 + 𝑦)3
c. (𝑥 + 𝑦)4
Notice that the amount of terms is always 1 greater than the exponent.
How do the coefficients of the simplified terms in your binomial expansions relate to Pascal’s
Triangle?
The coefficients related to the row of pascal’s.
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Binomial Theorem
 x  y
n
 n C0 x n  n C1 x n 1 y  n C2 x n  2 y 2  ...  n Ck x n k y k  ...  n Cn y n , where n  I , n  0
All binomial expressions will be written in descending order of the exponent of the first term in
the binomial.
The following are some important observations about the expansion of  x  y  , where x and y
n
represent the terms of the binomial and nεN:
 The expansion contains n + 1 terms
 The sum of the exponents in any term of the expansion is n.
Expand (𝑥 + 𝑦)4 using the binomial theorem:
 How many terms will we expect?

List the values of n, x and y:
x = x, n = 4, y = y

Expand using the formula:
 x  y
n
 n C0 x n  n C1 x n 1 y  n C2 x n  2 y 2  ...  n Ck x n k y k  ...  n Cn y n
4𝐶0 𝑥
Ex. Expand  3 x  2 
4 0
𝑦 + 4𝐶1 𝑥 3 𝑦1 + 4𝐶2 𝑥 3 𝑦 2 + 4𝐶3 𝑥 3 𝑦 3 + 4𝐶4 𝑥 0 𝑦 4
3
Ex. Expand (𝑥 − 7)5
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General Term of the Expansion of ( x  y)n
The term n Ck x n  k y k is called the general term of the expansion.
It is the  k  1 term in the expansion (not term k)
th
t  n C x n k y k
t 1
k
Note: expansion starts at a ‘k’ value of 0, so the ‘k’ value is one less than term requested
Ex. a) Find the fifth term of  x  y 
8
Fifth Term, think about the expansion:
Dealing with the ‘x’ values:
𝑘8 + 𝑘7 + 𝑘6 + 𝑘5 + 𝑘4
Notice after 5 terms our exponent is 4
Dealing with the ‘y’ values:
𝑘 0 + 𝑘1 + 𝑘 2 + 𝑘 3 + 𝑘 4
Notice after 5 terms our exponent is 4
9 terms, n=8, k = 4
tk 1  n Ck x n k y k
t41   8 C4   x 4  y 4 
t5  70 x 4 y 4
b. the middle term of  2 x  5 
6
7 terms, n = 6, k = 3 (middle term would be the 4th term)
c. Third term of (2𝑥 + 3)5
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Ex. One term in the expansion of  x  a  is 3 281 250 x 4 . Determine the
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numerical value of a.

What term will be 𝑥 4 ?
This ensures the x values will cancel out because they will have the same exponent.
n
Ck x nk y k  3281250 x 4
10
C6 x106 y 6  3281250 x 4
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1 15
Ex. Find the constant term (the term independent of x) in the expansion of (2𝑥 − 𝑥 2 )
To have a constant term, what should the exponent of ‘x’ be?
General Strategy:
1. Put equation into general term formula
2. Simplify/ expand formula
3. Solve for exponent of variables (gives k value)
Break down the equation to get x by itself:
Rewrite equation : (2𝑥 +
= 𝑛𝐶𝑘 𝑥
=
(−1) 15
) , n = 15, k = k
𝑥2
𝑛−𝑘 𝑘
𝑦
15−𝑘
(−
15𝐶𝑘 (2𝑥)
1 𝑘
)
𝑥2
1 𝑘
)
𝑥2
= ( 15𝐶𝑘 ) (−1)𝑘 (𝑥15−𝑘 )(215−𝑘 )(𝑥 −2 )𝑘
= ( 15𝐶𝑘 ) (−1)𝑘 (𝑥15−𝑘 )(𝑥 −2𝑘 )(215−𝑘 )
= ( 15𝐶𝑘 ) (−1)𝑘 (𝑥15−3𝑘 )(215−𝑘 )
(𝑥15−3𝑘 ) = 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑡ℎ𝑒 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑒𝑞𝑢𝑎𝑙 0
15 − 3𝑘 = 0
5 = 𝑘, 𝑡𝑒𝑙𝑙𝑠 𝑢𝑠 𝑡ℎ𝑒 𝑡𝑒𝑟𝑚 𝑛𝑢𝑚𝑏𝑒𝑟
Use general term formula:
1 5
10
= 15𝐶5 (2𝑥) (− 2 )
𝑥
1
= (3003)(1024𝑥10 ) (− 10 )
𝑥
= −3075072
= ( 15𝐶𝑘 ) (2)15−𝑘 (𝑥)15−𝑘 (−1)𝑘 (
Workbook: pg. 410 – 412 #4, 7-13 , pg. 415 #1abcdefg
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