Unformatted text preview: MATH 121 DISCUSSION WEEK 8 SOLUTIONS 1) 5 × 4 × 3 × 6 × 6 × 6 = 12,960 possible ﬂoral arrangements 2) P(5, 5) = 5! = 5! = 120 possible ways to stand in line 0! 3) P(6, 2) = 6! = 6 × 5 = 30 possible ways to pick 1st and 2nd chair 4! 4) P(10, 4) = 10! = 10 × 9 × 8 × 7 = 5,040 possible ways to allocate prizes 6! 8! 5) 3!3! = 1,120 possible arrangements 6! 6) 3!2! = 60 possible arrangements 8 7) C(8, 4) = 4!4! = 70 types of fruit salad 20! 8) C(20, 8) = 12!8! = 125,970 possible boards of directors 40! 9) P(40, 3) = 37! = 59,280 possible locker combinations 10! 10) C(10, 3) = 7!3! = 120 possible committees 10! 11) 2!2!3!3! = 25,200 possible arrangements 12) 36 × 36 × 26 × 26 × 10 × 10 × 10 = 362 × 262 × 103 = 876,096,000 possible passcodes 13) 26 × 25 × 24 × 10 × 9 × 8 = 11,232,000 possible passcodes 20! 14) C(20, 5) = 15!5! = 15,504 possible combinations 11! 15) 4!4!2! = 34,650 possible arrangements 16) P(9, 3) = 9! = 504 possible sets of oneway tickets. Order matters here because we are 6! talking about possible travel routes through 3 cities (so for example, suppose you start your trip in city A and go to B and C. This is not the same as starting in city B and then going to A and C, ie. A → B → C is not the same route as B → A → C ). 52! 17) C(52, 5) = 47!5! = 2,598,960 possible hands 5! 18) C(5, 3) = 2!3! = 10 possible subsets. Although our set has 10 elements total, not all of them are odd – only 5 of them are, and we are only going to pick 3 from those 5 (we don’t pick from the even numbers). 1 ...
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This note was uploaded on 12/07/2010 for the course MATH Math 121 taught by Professor Beaulieu during the Fall '10 term at UMass (Amherst).
 Fall '10
 Beaulieu
 Math, Probability

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