We know that the balls and buckets are distinguishable, and n>2. I was searching the site for some clues but didn't really find any of them helpful. Any thoughts?
2026-04-11 21:17:02.1775942222
Dividing n balls into n buckets so 2 are empty.
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I'm going to assume the question is, how many ways are there to distribute $n$ distnguishable balls among $n$ distinguishable buckets in such a way as to leave exactly two buckets empty.
First, choose the two buckets to be empty --- $n\choose2$ ways to do this.
Now you could have $3$ balls in one bucket, one ball in every other bucket. $n-2$ ways to choose the 3-ball bucket, $n\choose3$ ways to choose the balls to go in it, $(n-3)!$ ways to distribute the other balls.
Or, you could have $2$ balls in each of two buckets, one ball in each other bucket. $n-2\choose2$ ways to choose the two buckets, $n\choose4$ ways to choose the four balls that will go into those two buckets, $4\choose2$ ways to distribute the four balls to the two buckets, $(n-4)!$ ways to dispose of the other balls.
Can you combine all these numbers to get the answer?