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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
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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 ؍ 201600 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
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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 ؍ 117600 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
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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
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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)
؍ 36015x4
؍ 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

Analyses of adverse events during apheresis with focus on substitution fluids. The Swedish registryRut Norda1, Ulla Axdorph2, Claes-Göran Axelsson3, Gösta Berlin4, Svante Jonsson5, Gunilla Kumlien6, Jan Kurkus7, Bernd Stegmayr8, Björn Wikström91. Dept. of Clin. Imm. & Transfusion Medicine, Uppsala; 2. Dept. of Clin Imm. & Transfusion Medicine, Solna; 3. Dept. of Transfusion Medicine,

DISPENSING BY NON-PROPRIETARY NAME IS AUTHORIZED IF NOT CHECKED IN THIS COLUMN PHARMACOLOGIC TREATMENT OF DELIRIUM ORDER SET Pharmacy and Therapeutics Committee There is documentation of the diagnosis of delirium and the possible cause, if known, and any need for further diagnostic tests to meet the goals of care. If appropriate, the cause of the delirium is being treated. PRN medicat