donut distribution

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by Jeff@TargetTestPrep » Mon Dec 18, 2017 4:47 pm
vivekjaiswal wrote:Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
Let D be a donut, so we have DDDDD to distribute among three people. Also, we can use "|" as a separator, and we need two of them since there are three people. For example, D|DDD|D means Larry gets 1 donut, Michael 3, and Doug 1, and DDDDD|| means Larry gets 5 donuts, Michael 0, and Doug 0. Thus, the problem becomes how many ways we can arrange 5 Ds and 2 strokes. To solve it, we can use the formula for permutation of indistinguishable objects:

7!/(5! x 2!) = (7 x 6 x 5!)/(5! x 2) = 42/2 = 21

Answer: A

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