Exam2_20.110_review_solutions

# Exam2_20.110_review_solutions - Exam 2 Review 1 Solutions...

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Exam 2 Review 10/26/2015 1 Solutions 1. (a) Given one person, there are 5 choices for his/her partner. Given one the remain- ing 4 people, there are 3 choices for his/her partner. This leaves 2 more people who have to be partners. So the total number of ways 6 people can pair up is 5 × 3 × 1 = 15. Given a male, there are three choices for his partner. Then 2 remaining choices for the next male and 1 choice for the last male. So the total number of male-female partnerships is 3 × 2 × 1 = 6. (b) Out of 5 holes, we need to choose 2 to have a green ball. This gives ( 5 2 ) = 10. (c) Here are all the ways 1 green ball and 2 red balls can be placed in 3 holes: hole1 hole2 hole3 grr none none none grr none none none grr gr r none gr none r r gr none none gr r r none gr none r gr rr g none rr none g g rr none none rr g g none rr none g rr g r r r g r r r g (d) There are three ways to get all three balls in one hole. This corresponds to an energy level of 9. There are 12 ways to get 2 balls in one hole and 1 ball in another hole. This corresponds to an energy level of 4+1 = 5. There are 3 ways to one

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