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How many ways are there to distribute $k$ balls into $n$ distinct boxes ($k < n$) with at most one ball in any box if

(a) The balls are distinct?
(b) The balls are identical?

My initial response to this question's part (a) is $\binom n k$ but I have a small feeling this is incorrect, so I was wondering if someone could clear this first part up so I can start working on part (b) without having a false foundation to work with.

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  • $\begingroup$ That's the answer to (b): we choose the $k$ boxes that will be occupied. $\endgroup$ – André Nicolas Oct 28 '13 at 23:46
  • $\begingroup$ Dividing your answer in (a) by $k!$ should give you answer in (b) $\endgroup$ – Alex Oct 29 '13 at 0:47
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Solution to (a):

For the first ball you have $n$ choices, for the second $n-1$ and so on. So the total number of choices for $k$ balls is $(n-k+1)!$

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As a small exercise to sharpen your problem solving skills, I'd suggest you try it out for small $n, k$ on paper. Start with, say $n = 3, k = 2$ or $n = 4, k = 3$. You should quickly see how the dynamics work in both situations.

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a)Your problem reduces to choosing k out of n different boxes and then arranging them. This can be done in C(n,k).k! ways.

b)If the balls are identical, you just have to choose k out of n boxes. there are C(n,k) ways to do this.

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