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quiz2solutions - 16-2. Problem 2. For each integer between...

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Quiz 2 Solutions Dilip Raghavan October 7, 2008 Problem 1. (a) It is best to count the complement: the number of distributions where one of the 2 boxes is empty. First to count the total number of distributions, first distribute the red balls: ± 4 + 2 - 1 4 ² ways. Next distribute the blue balls: ± 5 + 2 - 1 5 ² ways. And finally the yellow balls: ± 7 + 2 - 1 7 ² ways. So there are a total of ³ 4 + 2 - 1 4 ! × ³ 5 + 2 - 1 5 ! × ³ 7 + 2 - 1 7 ! ways to distribute the balls without restrictions. Now if one of the boxes is empty, then all the balls must go into the other box. So there are 2 distributions with one box empty. So the answer is ³ 4 + 2 - 1 4 ! × ³ 5 + 2 - 1 5 ! × ³ 7 + 2 - 1 7 ! - 2 (b) Again count the complement. If the balls are all distinct, there are a total of 16 balls. The total number of distributions is then 2 16 (for each ball you have 2 choices, box 1 or box 2). There are again 2 distributions where all the balls end up in one of the boxes. So the answer is 2
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Unformatted text preview: 16-2. Problem 2. For each integer between 1 and 15 we have 2 choices: we can put it in box 1 or in box 2. Since box 1 has exactly ve integers we must use box 1 ve times and so, we must use box 2 ten times. So we are really arranging ve 1s and ten 2s in row. The requirement that consecutive integers cannot go into box 1 means that there are no consecutive 1s in the arrangement. So let us rst arrange the ve 1s: 1 1 1 1 1 Now we want to distribute the 2s making sure that there are no consecutive 1s. So rst put a 2 into each of the 4 spaces between the 1s. Now, we are left with 6 2s to distribute into the 6 spaces. So there are 6 + 6-1 6 ! ways to do this. 1...
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