Randomly, $k$ distinguishable balls are placed into $n$ distinguishable boxes, with all possibilities equally likely. Find the expected number of empty boxes.
PROPOSED SOLUTION:
Let $I_j$ be the indicator random variable for the $j^{th}$ box being empty, so $I_1 + ··· + I_n$ is the number of empty boxes.
Now $E(I_j) = P(I_j = 1) $
Given both boxes AND balls are distinguishable total number of ways to place balls in boxes is $(n + k)!$
Now lets assume $j^{th}$ box is left empty, remaining $(n-1)$ boxes can be filled as $(k + n - 1)!$ ways
Therefore, $I_j = \dfrac{(k + n -1)!}{(k + n)! }= E(I_j)$
By linearity of Expectation, $E(\sum_{k=1}^n[I_j]) = \sum_{k=1}^n(E(I_j))$
= $\sum_{k=1}^n(\dfrac{(k + n -1)!}{(k + n)!})$
= $\dfrac{n(k + n - 1)!}{(k + n)!}$
solution guide gives answer as $n(1 - 1/n)^k$ which is arrived at by saying $I_j = (1 - 1/n)^k$
Any insights in what is wrong above.
Imagine tossing the balls one at a time into bins. For each ball, there are $n$ equally likely places for it to land, for a total of $n^k$ equally likely cases. This is quite different from your $(n+k)!$.
For any fixed bin B, the probability any individual ball falls outside B is $\frac{n-1}{n}$. Thus the probability they all land outside B is $\left(\frac{n-1}{n}\right)^k$.