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I got this question which i couldn't figure out:

Given $m,n,k \geq 1$ such that $k \ge n-m$. in how many ways can we put $k$ diffrent balls in $n$ boxes, that excatly $m$ boxes will be empty?

i thought about $n \choose m$ for the choosing the empty boxes, and ${n-m-k-1} \choose {n-m}$ for putting the balls in the remainig boxes. The result will be the multiplication of these two.

Am i right?

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    Looks about right. How many different ways can you choose the $n-m$ boxes that must be filled? Then, in each such case, every one of the $n-m$ boxes must have $1$ ball, leaving $k-(n-m)$ balls that you can distribute any which way you like. So how many different ways can you distribute $k-(n-m)$ balls in $n-m$ boxes? That's the product you want. – 2017-01-10
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    It says $k$ *different* balls. What about the boxes ? Identical boxes or not ? – 2017-01-10
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    both boxes and balls are diffrent. i thought about \binom{n}{m}*\binom{n−m−k−1}{n−m} and after some algebra we get {n!\over (m!*((k-1)!)^2)} what about that ? – 2017-01-11

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