Ways to distribute $25$ distinct balls to $40$ bins with constraints

Question

How many ways there are to distribute $25$ distinct balls to $40$ bins such that:

1. At least one bin will have more than one ball.

2. One and only one bin (not predetermined) will hold exactly 10 balls. (In other words, only one bin will hold 10 balls and all others will have less or more than 10)

My attempt:

1. with complement, the general case: each ball has 40 bins to go to $25^{40}$.

   At most one ball in every bin: $\binom {40} {25} 25!$

   Total: $25^{40}-\binom {40} {25} 25!$

2. Choose one bin: $\binom {40} 1=40$.

   choose $10$ balls: $\binom {25} {10}$ and place them in the above bin: $40 \binom {25}{10}$

   Now there are $39$ bins and $15$ balls left, we want that for all, bins none has 10 balls, so the complement would be that there's one bin with 10 balls:

   The general case: $15^{39}$. Choose the other bin to hold 10 balls: $39$, choosing the balls that would go to that bin: $\binom {15} {10}$, $5$ balls left and $38$ bins: $5^{38}$

   Total: $(40 \binom {25}{10})(15^{39}-39\binom {15} {10}5^{38})$

Answer 1

Community wiki answer so the question can be marked as answered:

As noted in comments, your calculation is basically right; the only error is that wherever you count the number of ways to put $k$ distinct objects into $n$ bins you have $k^n$ instead of $n^k$.