<s>[INST] Distribution of different objects into different boxes We want to put n different objects into n different boxes. In how many ways can we do this if we want that exactly two boxes remain empty?
 [/INST] The number of ways to choose the two boxes that are to remain empty is a combination:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$
As for the remaining boxes, there can either be $1$ box with $3$ objects and $n-3$ boxes with $1$ object each, or $2$ boxes with $2$ objects each and $n-4$ boxes with $1$ object each.

Case 1: One box has three objects.
The number of ways to choose this special box is $n-2$. The number of ways to choose the three objects that go in this one box is $$\binom{n}{3}$$. Then, the number of ways to permute the remaining $n-3$ objects in the $n-3$ boxes is simply $(n-3)!$. Thus, the total number of ways to place the objects in this case is $$\binom{n}{2} \cdot (n-2) \cdot \binom{n}{3} \cdot (n-3)!$$

Case 2: Two boxes have two objects.
The number of ways to choose the two special boxes is $$\binom{n-2}{2}$$ Once these boxes are chosen, the number of ways to choose the objects that go in the first of these is $$\binom{n}{2}$$ After choosing these, the number of ways to choose the objects that go in the second box with two things in it, since there are $n-2$ objects left to choose from, is $$\binom{n-2}{2}$$ Finally, the number of ways to permute the remaining $n-4$ objects in the remaining $n-4$ boxes is $(n-4)!$. This gives the number of placements in this case as $$ \binom{n-2}{2} \cdot \binom{n}{2} \cdot \binom{n-2}{2} \cdot (n-4)!$$

The final answer is the sum of the answers in cases 1 and 2, which is
$$\boxed{\binom{n}{2} \cdot (n-2) \cdot \binom{n}{3} \cdot (n-3)! + \binom{n-2}{2} \cdot \binom{n}{2} \cdot \binom{n-2}{2} \cdot (n-4)!}$$

Edit: I noticed that the final answer, as I wrote it, could be simplified quite nicely. First, we can expand the binomial coefficients and combine $n-2$ and $(n-3)!$:
$$\frac{n(n-1)}{2} \cdot (n-2)! \cdot \frac{n(n-1)(n-2)}{6} + \frac{(n-2)(n-3)}{2} \cdot \frac{n(n-1)}{2} \cdot \frac{(n-2)(n-3)}{2} \cdot (n-4)!$$
Then, we find that both terms have $n!$ in them:
$$\frac{n!}{2}\binom{n}{3} + \frac{n!}{4}\binom{n-2}{2}$$
$$\frac{n!}{4}\cdot\frac{n(n-2)(n-3)}{3} + \frac{n!}{4}\cdot\frac{(n-2)(n-3)}{2}$$
This ends up as the relatively concise expression
$$\frac{n!(n-2)(n-3)}{4}\left(\frac{n}{3}+\frac{1}{2}\right)$$
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