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10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 32 R E V I E W, p a g e s 7 5 4 – 7 5 8
1. A penny, a dime, and a loonie are in one bag. A nickel, a quarter, and
a toonie are in another bag. Tessa removes 1 coin from each bag. Usea graphic organizer to list the total amounts she could haveremoved.
Use an organized list.
1¢ ؉ 5¢ ؍ 6¢
10¢ ؉ 5¢ ؍ 15¢
$1 ؉ 5¢ ؍ $1.05
1¢ ؉ 25¢ ؍ 26¢
10¢ ؉ 25¢ ؍ 35¢
$1 ؉ 25¢ ؍ $1.25
1¢ ؉ $2 ؍ $2.01
10¢ ؉ $2 ؍ $2.10
$1 ؉ $2 ؍ $3
So, there are 9 total amounts that Tessa could have removed.
2. The Braille code consists of patterns of raised dots arranged
in a 3 by 2 array. The pattern for the letter Z is shown.
How many different patterns are possible? Explain your reasoning.
Each position in the 3 by 2 cell may or may not have a raised dot.
Use the fundamental counting principle.
The number of possible patterns is: 2 # 2 # 2 # 2 # 2 # 2 ؍ 64
3. A hiking group consists of 12 students and 2 leaders. A leader must
be at the front and back of the line. How many ways can the grouphike in a line? The number of ways the students can hike in a line is:
12! ؍ 479 001 600
The leaders A and B must be at the front and back of the line.
There are 2 ways for the leaders to line up: AB and BA
So, the number of ways the group can hike in a line is:
479 001 600 # 2 ؍ 958 003 200
Chapter 8: Permutations and Combinations—Review—Solutions 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 33 4. A code consists of 4 letters from the English alphabet and 3 letters
from the Greek alphabet. There are 8 English letters and 6 Greekletters to choose from and repetition is not allowed. How many 7-letter codes are possible? Order matters, so use permutations.
Determine the number of
Determine the number of
ways to arrange 4 English letters
ways to arrange 3 Greek letters
chosen from 8 letters:
chosen from 6 letters:
8 # 7 # 6 # 5 ؍ 1680
6 # 5 # 4 ؍ 120
Use the fundamental counting principle.
1680 # 120 ؍ 201 600
So, there are 201 600 possible 7-letter codes.
5. How many 12-letter permutations of GOBBLEDEGOOK can be
There are 12 letters: 2 are Gs, 2 are Es, 2 are Bs, and 3 are Os.
Number of permutations:
2!2!2!3!
So, 9 979 200 permutations can be created.
6. How many ways are there to arrange all the words in this tongue
twister? CAN YOU CAN A CAN AS A CANNER CAN CAN A CAN? There are 12 words: 6 are CAN, and 3 are A.
Number of permutations:
12! ؍ 110
So, there are 110 880 ways to arrange all the words.
7. How many 9-letter permutations of EQUATIONS can be created if
the vowels must appear together in the order A, E, I, O, and U?Explain.
There are 5 vowels and 4 consonants.
Since the vowels have to be together, consider them as 1 object.
So, there are 5 objects: the group of vowels and 4 different consonants.
The number of permutations of 5 objects is: 5! ؍ 120
The vowels must be in the given order, so the number of permutations of
the vowels is 1.
So, all the letters can be arranged in 120 ways.
Chapter 8: Permutations and Combinations—Review—Solutions 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 34 8. A student volunteers at a food bank. He fills hampers with these items:
5 cans of soup chosen from 7 different soups3 bags of pasta chosen from 4 different types of pasta4 bags of vegetables chosen from 8 different types of vegetables3 boxes of cereal chosen from 6 different types of cerealHow many ways can the student fill a hamper? Use combinations. The number of ways of choosing:
5 cans of soup from 7 different soups is: C ؍ 21
3 bags of pasta from 4 different types of pasta is: C ؍ 4
4 bags of vegetables from 8 different types of vegetables is: C ؍ 70
3 boxes of cereal from 6 different types of cereal is C ؍ 20
Use the fundamental counting principle.
21 # 4 # 70 # 20 ؍ 117 600
The student can fill the hamper in 117 600 ways.
9. Solve each equation for n or r.
(n ؊ 3)!3!
(7 ؊ r)!r!
(n ؊ 3)!6
(7 ؊ r)!r!
6 # 84 ؍ n(n ؊ 1)(n ؊ 2)
(7 ؊ r)!r! ؍ 5040
504 ؍ n(n ؊ 1)(n ؊ 2)
Use a calculator.
(7 ؊ r)!r! ؍ 144
3 504 Џ 7.96
Use guess and test.
So, try 3 consecutive
Since 4! ؍ 24,
numbers with 8 as the
(7 ؊ 4)!4! ؍ 6 # 24
middle number:
7 # 8 # 9 ؍ 504
Or, since ,
10. a) These are the first 7 numbers in row 14 of Pascal’s triangle:
1, 13, 78, 286, 715, 1287, 1716Complete the row. What strategy did you use? Row 14 of Pascal’s triangle has 14 terms.
The triangle is symmetrical: the numbers in each row read the same
from left to right and from right to left. So, the remaining numbers in
row 14 are the given numbers in reverse order: 1716, 1287, 715, 286,
78, 13, 1
Chapter 8: Permutations and Combinations—Review—Solutions 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 35 b) Use your results from part a to write the numbers in row 15 of
1 1 ؉ 13 ؍ 14
13 ؉ 78 ؍ 91
78 ؉ 286 ؍ 364
286 ؉ 715 ؍ 1001
715 ؉ 1287 ؍ 2002 1287 ؉ 1716 ؍ 3003
1716 ؉ 1716 ؍ 3432 1716 ؉ 1287 ؍ 3003 1287 ؉ 715 ؍ 2002
715 ؉ 286 ؍ 1001
286 ؉ 78 ؍ 364
78 ؉ 13 ؍ 91
13 ؉ 1 ؍ 14
So, the numbers in row 15 are:
1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1
11. Determine the value of each number in Pascal’s triangle.
a) the 4th number in row 13
C Substitute: n
؍ 12, r ؍ 3
b) the 5th number in row 21
C Substitute: n
؍ 20, r ؍ 4
12. The 4th term in the expansion of (x + 1)8 is 56x5. Which other term
in the expansion has a coefficient of 56? Explain.
Because both coefficients in the binomial (x ؉ 1)8 are 1, the coefficients in
the expansion are the 9 terms in row 9 of Pascal’s triangle. These terms are
the same when read from left to right and right to left. So, in row 9, the 5th
term is the middle term, and the 6th term is the same as the 4th term, 56.
13. Expand using the binomial theorem.
a) (4x - 1)5
(x ؉ y)n ؍ C xn ؉ C xn؊1y ؉ C xn؊2y2 ؉ . . . ؉ C yn
Substitute: n ؍ 5, x ؍ 4x, y ؍ ؊1
(4x ؊ 1)5 ؍ C (4x)5 ؉ C (4x)4(؊1) ؉ C (4x)3(؊1)2 ؉ C (4x)2(؊1)3
؉ C (4x)(؊1)4 ؉ C (؊1)5
؍ 1(1024x5) ؉ 5(256x4)(؊1) ؉ 10(64x3)(1) ؉ 10(16x2)(؊1)
؉ 5(4x)(1) ؊ 1
؍ 1024x5 ؊ 1280x4 ؉ 640x3 ؊ 160x2 ؉ 20x ؊ 1
Chapter 8: Permutations and Combinations—Review—Solutions 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 36 b) (5x2 + 2y)4
(x ؉ y)n ؍ C xn ؉ C xn؊1y ؉ C xn؊2y2 ؉ . . . ؉ C yn
Substitute:
(5x2 ؉ 2y)4 ؍ C (5x2)4 ؉ C (5x2)3(2y) ؉ C (5x2)2(2y)2
؉ C (5x2)(2y)3 ؉ C (2y)4
؍ 1(625x8) ؉ 4(125x6)(2y) ؉ 6(25x4)(4y2) ؉ 4(5x2)(8y3) ؉ 16y4
؍ 625x8 ؉ 1000x6y ؉ 600x4y2 ؉ 160x2y3 ؉ 16y4
14. Determine the indicated term in each expansion.
a) the 3rd term in (-7x + 1)6
b) the 7th term in (6x + 2y)7
The kth term is: C
xn؊(k؊1)yk؊1
The kth term is: C
xn؊(k؊1)yk؊1
Substitute: k ؍ 3, n ؍ 6,
Substitute: k ؍ 7, n ؍ 7, x ؍ 6x,
x ؍ ؊7x, y ؍ 1
C ( ؊7x)4(1)2 ؍ 15(2401x4)(1)
C (6x)1(2y)6 ؍ 7(6x)(64y6)
؍ 36 015x4
؍ 2688xy6
The 3rd term is 36 015x4.
The 7th term is 2688xy6.
15. Use the binomial theorem to evaluate (1.2)6. Use a calculator to
Write (1.2)6 as .
(1 ؉ 0.2)6
(x ؉ y)n ؍ C xn ؉ C xn؊1y ؉ C xn؊2y2 ؉ . . . ؉ C yn
Substitute: n ؍ 6, x ؍ 1, y ؍ 0.2
(1 ؉ 0.2)6 ؍ C (1)6 ؉ C (1)5(0.2) ؉ C (1)4(0.2)2 ؉ C (1)3(0.2)3
؉ C (1)2(0.2)4 ؉ C (1)(0.2)5 ؉ C (0.2)6
؍ 1(1) ؉ 6(1)(0.2) ؉ 15(1)(0.2)2 ؉ 20(1)(0.2)3 ؉ 15(1)(0.2)4
؉ 6(1)(0.2)5 ؉ 1(0.2)6
؍ 1 ؉ 1.2 ؉ 0.6 ؉ 0.16 ؉ 0.024 ؉ 0.001 92 ؉ 0.000 064
؍ 2.985 984
Chapter 8: Permutations and Combinations—Review—Solutions

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