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Harvard Statistics 110: see #30, p. 30 of pdf

Randomly, k distinguishable balls are placed into n distinguishable boxes, with all possibilities equally likely. Find the expected number of empty boxes.

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I'm not convinced by the solution because, as illustrated by the picture of balls and boxes, there are $n^k$ different ways of filling the boxes by the balls, and there are $k$ different ways of one box being filled. One box can be filled by $1, 2, \dots k$ balls. Then, $P(I_j=0)=k/n^k$, and $P(I_j=1)=1-k/n^k$. But this will give weird results for $n=9, k=13$. I don't know but I feel there's something missing. May someone help?

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The flaw in your reasoning is that there are many more than $k$ ways for a particular box to be filled.

Take the concrete example of $n=2$ boxes and $k=3$ balls. You are correct that there are $n^k=8$ different ways to fill the boxes. These $8$ ways can be enumerated by specifying the box where ball 1 lands, the box where ball 2 lands, and the box where ball 3 lands as a vector of three slots. The possibilities are: $$(1,1,1), (2,1,1), (1,2,1), (1,1,2), (2,2,1), (2,1,2), (1,2,2), (2,2,2).$$ Read $(1,1,1)$ as: "ball 1 lands in box 1, ball 2 lands in box 1, ball 3 lands in box 1". Let's focus on how box 1 can be filled or not filled. By inspection we see that there are $7$ ways for box 1 to be filled, not $3$, and in only one of these possibilities is box 1 empty. Moreover, $1 = (2-1)^3=(n-1)^k$; that is, the only way box 1 can be empty is if each ball lands in a box other than 1, which means each slot in the $3$-vector can be filled only with one of $n-1$ choices.

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Consider there are k balls marked 1,2,3,4,.....k

and boxes marked as 1,2.....n

So to find the ways in which balls can be filled can be calculated as

  1. When box 1 is filled with 1 ball. The no of ways in which this can happen is by choosing 1 ball out of k balls and then each remaining k-1 balls will have n-1 choices to go into n-1 boxes = $ \binom{k}{1}*(n-1)^{k-1}$

  2. When box 1 is filled with 2 balls. The no of ways in which this can happen is by choosing 2 ball out of k balls and then each remaining k-2 balls will have n-1 choices to go into n-1 boxes = $ \binom{k}{2}*(n-1)^{k-2}$ ......... and so on.

  3. till $ \binom{k}{k}*(n-1)^{k-k}$

Total no of ways to be filled = $\binom{k}{1}*(n-1)^{k-1} + \binom{k}{2}*(n-1)^{k-2} ..... + \binom{k}{k}*(n-1)^{k-k}$

$ = [\binom{k}{1}*(n-1)^{k-1} + \binom{k}{2}*(n-1)^{k-2} ..... + \binom{k}{k}*(n-1)^{k-k} + \binom{k}{0}*(n-1)^{k-0}] - \binom{k}{0}*(n-1)^{k-0}$

$ = (1+(n-1))^k - \binom{k}{0}*(n-1)^{k-0}$

$ = n^k - (n-1)^k$

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