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elemprob-fall2010-page13

elemprob-fall2010-page13 - them in 3 boxes Let us make...

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having 4 people out of the 10. Any such committee will either contain A or will not. 9 3 is the number of committees that contain A and 3 out of the remaining 9 people, while 9 4 is the number of committee that do not contain A and contain 4 out of the remaining 4 people. Suppose one has 9 people and one wants to divide them into one committee of 3, one of 4, and a last of 2. There are 9 3 ways of choosing the first committee. Once that is done, there are 6 people left and there are 6 4 ways of choosing the second committee. Once that is done, the remainder must go in the third committee. So the answer is 9! 3!6! 6! 4!2! = 9! 3!4!2! . In general, to divide n objects into one group of n 1 , one group of n 2 , . . . , and a k th group of n k , where n = n 1 + · · · + n k , the answer is n ! n 1 ! n 2 ! · · · n k ! . These are known as multinomial coefficients. Suppose one has 8 indistinguishable balls. How many ways can one put
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Unformatted text preview: them in 3 boxes? Let us make sequences of o ’s and | ’s; any such sequence that has | at each side, 2 other | ’s, and 8 o ’s represents a way of arranging balls into boxes. For example, if one has | o o | o o o | o o o | , this would represent 2 balls in the first box, 3 in the second, and 3 in the third. Altogether there are 8 + 4 symbols, the first is a | as is the last. so there are 10 symbols that can be either | or o . Also, 8 of them must be o . How many ways out of 10 spaces can one pick 8 of them into which to put a o ? The answer is ± 10 8 ² . An example. How many quintuples ( x 1 ,x 2 ,x 3 ,x 4 ,x 5 ) of nonnegative in-tegers whose sum is 20 are there? Answer: This is the same as asking: how 13...
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