Download Chapter 11: Permutations, Combinations and Binomial Theorem

Chapter 11: Permutations, Combinations and Binomial Theorem
Lessons 1 and 2: Permutations
Specific Outcome 1: Apply the fundamental counting principle to solve problems.
Specific Outcome 2: Determine the number of permutations of n elements taken r at a time to solve problems.
Fundamental Counting Principle:
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
Determine how many different coloured toys can be produced:
3
top
.
2
.
4
middle
bottom
= 24
coloured toys
Check your answer using a tree diagram (if they don’t know how to make one, show
them)
Make a conjecture about how you can use multiplication only to arrive at the number of
different coloured toys possible.
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.
Another thing that you will need to consider is the idea of and vs. or. If the problem has
“and” in it, that means you are to multiply. If the problem has “or” or “at least” or “at
most”, it is talking about multiple situations and you need to add.
Ex. Determine the number of distinguishable four letter arrangements that
can be formed from the word ENGLISH if:
a. letters can be repeated? 7 . 7 . 7 . 7 = 2401
b. no letters are repeated and:
i)
there are no further restrictions?
7 . 6 . 5 . 4 = 840
ii)
the first letter must be E?
E
the “word” must contain G?
iii)
1 . 6 . 5 . 4 = 120
1 . 6 . 5 . 4 = 120
(120)(4) – the ‘G’ can be in any of the four slots
= 480
iv)
the first and last letters must be vowels? 2_. 5 .
V
4.
1 = 40
V
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?
1
.
1 . 1
.
10
.
10 . 10
1000
or
+
1 . 5
. 10 . 10 . 10 =
5000
= 6000 different telephone numbers
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?
24 . 9
.
10
.
10
.
26
.
25
= 14 040 000
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?
6 . 5 . 4 = 120
b) How many of these are:
i) less than 400?
2 . 5 . 4 = 40
ii) even?
5 . 4 . 2 = 40
( only 5 choices because between the 2
and the 6, only one will be available for the first number because one of them is
needed for the last, to be even)
2
iii) multiples of 5? 5 . 4 . 1 = 20 ( the last digit can only be 5, so that leaves 5
choices for the first digit)
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)(n3)….(3)(2)(1), where n W
Ex. Use the factorial key on your calculator to solve 6!
On Calc: Math – PRB – 4 - !
= 720
10!
, there are several approaches:
7!
a.Use you calculator: = 720
b. By Cancellation:
10! 10  9  8  7!

 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 720
7!
7!
7x6x5x4x3x2x1
Ex. To simplify
Ex. Find the value of
43! (43)(42)(41)(40!)

 74046
40!
40!
Ex. Simplify the following expressions:
a.
b.
n!
n(n  1)(n  2)!

 n(n  1)  n 2  n
(n  2)!
(n  2)!
 n  3 !
n!

(n  3)(n  2)(n  1)n !
 (n  3)(n  2)(n  1)
n!
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: 6! = 720 (6 letters)
b. KELOWNA: 7! = 5040 (7 Letters)
3
The number of permutations of “n” different objects taken “r” at a time is:
n!
n Pr  (n  r )!
(no repeats are allowed in this formula)
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!
8!
8! 8  7  6  5!

 
 8  7  6  336
n pr 
(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!
n!

n pn  n ! 
(n  n)! 0!
For this to be equal to n! , the value of 0! must be 1. Therefore 0! Is defined to have a
value of 1
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:
a) fundamental counting principle:
26 . 25 . 10 . 9 . 8 . 7 = 3 276 000
b) permutations:
(letters)(digits)
 26 P2  10 P4   (650)(5040)  3276000
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)!
4
n!(n-3)! = 28(n-1)!(n-4)! (cross multiplied)
n(n-1)!(n-3)! = 28(n-1)!(n-4)!
(n is expanded to n(n-1)! to simplify)
n(n-3)! = 28(n-4)! (cancel n-1 on each side)
n(n-3)(n-4)! = 28(n-4)! (expand n(n-3)!)
n(n-3) = 28 (cancel n-4’s)
n 2  3n  28  0
(n-7)(n+4) = 0
n=7 and n = -4 (extraneous – doesn’t fit equation)
In many cases involving simple permutations, the fundamental counting principle can be
used in place of the permutation formulas.
Permutations With Restrictions And Repetition
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: 7! = 5040
b.)the first letter must be an N? : 1 . 6 . 5 . 4 . 3 . 2 . 1 = 720
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.
The letters must be considered as a group of 1. Therefore, you are arranging
that one group plus 4 other letters. So 5!. However, there you need to arrange
the three letters themselves which is 3!. So (5!)(3!) = 720
b) the letters IT cannot be together:
It is easier to solve these questions if you consider the opposite situation Must Be
together. That way, you can take the total with no restrictions and then subtract
out the Must Be Together. The leftovers will be the must not be together answer.
5
Total no restrictions: 7! = 5040
Must be together: IT = 1 + 5 other letters = 6! X 2! (arrangement of IT) = 1440
Answer: 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?
4 . 3 . 3 . 2 . 2 . 1 . 1 = 144
B G B G B G B
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 !
**basically remember to divide out the repeats in factorial notation**
Ex. Find the number of permutations of the letters of the word:
9!
 181440
a.) VANCOUVER: there are 2V’s:
2!
b.) MATHEMATICAL: 2M’s, 3A’s, 2T’s:
12!
 19958400
(2!)(3!)(2!)
Ex. How many arrangements of the word POPPIES can be made under
each of the following conditions?
7!
 840
a.) without restrictions:
3!
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)
6!
=  360
2!
6
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, 1box 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)
=
1 9!  15120
 3! 2! 2!
One thing that you may see on the diploma are what we call pathway questions. They
can be done by using repeated permutations or drawing.
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
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)
4N 5W =
9!
= 126 ways (also show them how to draw these)
4!5!
Assignment: pg. 524-527 #1 – 8, 10 – 20, 22 – 25, 27 – 28, 32, C3a, C5 (pick one
letter from each question)
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Lesson 3: Combinations
Specific Outcome 3: Determine the number of combinations of n different elements taken r at a time to solve
problems.
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
2nd position
3rd position
President
Vice President
Treasurer
a. Does the order in which the students are elected matter? Why?
Yes. The vice president is not the president (and vice versa). They have distinct
roles.
b. In how many ways can the positions be filled from this group?
4 x 3 x 2 = 24
2. Now suppose that the three students are to be selected to serve on a committee.
a. Is the order in which the three students are selected still important? Why or
why not?
Not important, there are no distinct titles.
b. How many possible committees from the 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.
a. How many handshakes are possible if each student shakes every other
student’s hand once?
5 + 4 + 3 + 2 + 1 = 15
8
4. What is the biggest difference between a permutation and a combination?
A combination is a selection of a group of objects, taken from a larger group for which
the kind 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.
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?
10 C3  120
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?
49
C6  13983816
9
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?
6
C3  20
b. In how many ways can the males be chosen?
9
C4  126
c. In how many ways can the sub-committee be chosen?
Exactly 3 females
(females)(Males)
  6 C3  9 C4 
  20 126   2520
d. In how many ways can the sub-committee be chosen if Bruce the football
coach must be included?
(females)(males)(Bruce)
 6 C3  8 C3  1C1 
  20  56 1  1120
Combinations which are equivalent
Ex. Jane calculated 10 C2 to be 45 arrangements. She then calculated
be 45 arrangements. Use factorial notation to prove this:
10
C2  10 C8
10!
10!
2!8!
8!2!
10  9  8! 10  9  8!
2!8!
2!8!
45
=
45
10
10
C8 to
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
n(n  1)(n  2)!
 600
(n  2)!
n(n  1)  600  0
C2  300
n!
 300
(n  2)!2!
n!
 600
(n  2)!
n 2  n  600  0
(n  25)(n  24)  0
n  25, 24(extraneous)
25 players attended the tryouts
Homework: pg. 534 #1 – 6, 7a, 8 – 20, 22, 23, C1, C3 (pick one letter from each
question) #12 is a challenge
11
Lesson 4: The Binomial Theorem
Specific Outcome 4: Expand powers of a binomial in a variety of ways, including using the binomial theorem
(restricted to exponents that are natural numbers).
Pascal’s Triangle:
Show the first couple rows of the Pascal’s triangle. See if the students can find the
pattern and continue the next few rows.
Ex. Complete the following expansions:
a. (𝑥 + 𝑦)2
𝑥 2 + 2𝑥𝑦 + 𝑦 2
b. (𝑥 + 𝑦)3
𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3
c. (𝑥 + 𝑦)4
𝑥 4 + 4𝑥 3 𝑦 + 6𝑥 2 𝑦 2 + 4𝑥𝑦 3 + 𝑦 4
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.
Pascal’s Triangle can also be represented using combinations:
1
2
C0
0
C0
C0
1
2
C1
C1
2
C2
C0 3 C1 3 C2 3 C3
4 C1
4 C0
4 C2
4 C3
4 C4
3
These values are identical to the values in Pascal’s triangle above.
Notice these results from Pascal’s Triangle:
 the sum of the numbers in the kth row of Pascal’s triangle is 2k 1

take the 3rd row, add it up (equals 4), 231  22  4
n
the sum of the coefficients in the expansion of  x  y  is 2 n
( x  y)3  x3  3x 2 y  3xy 2  y 3 (notice the number of terms (4))
coefficients = 1 + 3 + 3 + 1 = 8 ( 23  8 )
12
Notice these results from combinatorics:


n
n C0  n C1  n C2  ...  n Cn1  n Cn  2
6
C0  6 C1  6 C2  6 C3  6 C4  6 C5  6 C6  26  64
1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 (pascal’s 7th row)
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
n
expansion of  x  y  , where x and y 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.
Ex. Expand  3 x  2 
3
4 terms, n = 3, x = 3x, y = (-2)
n
C0 x 3  n C1 x n 1 y  n C2 x n  2 y 2  n C3 x n 3 y 3
 3 C0  3 x   3 C1  3x   2   3 C2  3x   2   3 C3  2 
3
2
1
2
3
 1  27 x 3    3  9 x 2   2    3  3 x  4   1 8 
 27 x 3  54 x 2  36 x  8
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
13
Ex. a) Find the fifth term of  x  y 
8
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)
t31  6 C3  2 x   5 
3
3
 20 8 x3   125 
t4  20000 x3
Ex. One term in the expansion of  x  a 
value of a.
10
is 3 281 250 x 4 . Determine the numerical
n = 10 , k = (10 – 4(exponent) = 6) this gives us the same x value power, the position of
𝑥 4 in the expansion. The k value of 6 means the 7th term (7 – 1 = 6)
𝑦 0 + 𝑦1 + 𝑦 2 + 𝑦 3 + 𝑦 4 + 𝑦 5 + 𝑦 6 (this represents the k value of 6 for the y in
(𝑥 + 𝑦)𝑛 , 𝑛𝑜𝑡𝑖𝑐𝑒 𝑖𝑡 𝑖𝑠 7 𝑡𝑒𝑟𝑚𝑠)
𝑥10 + 𝑥 9 + 𝑥 8 + 𝑥 7 + 𝑥 6 + 𝑥 5 + 𝑥 4 (expansion for the x in (𝑥 + 𝑦)𝑛 , 𝑛𝑜𝑡𝑖𝑐𝑒 𝑖𝑡 𝑖𝑠 7 𝑡𝑒𝑟𝑚𝑠)
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
210 x 4 y 6  3281250 x 4
210 y 6  3281250
y 6  15625
a y 5
14
1 15
Ex. Find the constant term (the term independent of x) in the expansion of (2𝑥 − 𝑥 2 )
General Strategy:
1. Put equation into general term formula
2. Simplify/ expand formula
3. Solve for exponent of variables (gives k value)
Rewrite equation : (2𝑥 +
(−1) 15
𝑥2
) , n = 15, k = k
= 𝑛𝐶𝑘 𝑥 𝑛−𝑘 𝑦 𝑘
=
15𝐶𝑘
(2𝑥)15−𝑘
1 𝑘
(− 2 )
𝑥
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 (
Assignment p. 542 #1 – 11, 13 – 20, 22 – 23, C2 (pick one letter from each question)
15