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Unformatted text preview: HW2 Solutions Dilip Raghavan September 13, 2008 1 Section 5.2 Problem 28. Though the problem does not make this clear, I’m assuming that the teams are not ordered. That is, if I take all the members of Team 1 and swap them with all the members of Team 2, then that counts as the same outcome. Let us first count the outcomes as if the teams were ordered. Then in the end we can divide the answer by two. Count the complement – the number of ways to split into two teams where all the women end up on one of the teams. Let us first count all ways to split into two teams without any restrictions. Note that once the members of Team 1 are determined, Team 2 are also determined. So it is enough to choose the members of Team 1. There are 8 4 ways to do this. Now if all the women end up on one of the two teams, we must first decide which team will the contain all the women. There are 2 ways to do this (either Team 1 or Team 2). Once this has been decided we have choose one man for that team. There are 5 ways to do this. This determines all the members of that team. But then the other is also determined (they are the ones left over). So there are 2 × 5 ways to split into two teams so that all the women end up on the same team. What we are looking for is the complement: 8 4 2 × 5. Finally, since we are assuming the teams are not ordered, we divide the answer by 2. So 1 2 h 8 4 2 × 5 i is the answer. Problem 43. The total number of five card hands is 52 5 . (a) There are two cases: either all five cards are among Ace, King, Queen and Jack, or there is one other value. • Case I: all cards are among the four values. In this case, there are 2 cards of the same value and one from each of the other three. So first choose the value with 2 cards. There are 4 ways to do this. Next choose 2 cards of that value: 4 2 ways. Now choose one from each of the other three values: 4 1 3 . So the total is 4 × 4 2 × 4 1 3 . • Case II: there is one other value. Choose that value. There are 13 4 ways to do this. Next choose one card from each of the 5 values: 4 1 5 ways. So total is (13 4) × 4 1 5 . The total number of favorable outcomes is the sum of the two cases: 4 × 4 2 × 4 1 3 + (13 4) × 4 1 5 (b) There are 3 cases: no hearts and no spades, one heart and one spade and two hearts and two spades....
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This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.
 Spring '10
 DR.RAHULA

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