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In a movie theater there are $n$ rows and $k_1,k_2,...,k_n$ chairs in each row. How many different ways can $m$ people sit in this movie theater?

My thought process: There are $k_1+\dots+k_n$ places to sit. The first person has $k_1+\dots+k_n$ many choices, the second has $k_1+\dots+k_n-1$ many choices and the $m.$ person has $k_1+\dots+k_n-m+1$ many choices. So there are ${k_1+\dots+k_n\choose m}m!$ many ways. Is that correct?

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    Is it not just $m$ out of $\sum k_i$? – 2017-02-16
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    It depends on what we care about. If we care about which person sits in what seat, then you're right. If we only care about what seats are taken, but not who sits where, then the comment above tells you how to do that. We could care about what person sits in what row, but not exactly what seat. Or we could care about how many people sit in each row, but not necessarily which seat. As you can see, there are many interpretations. It all comes down to what "different ways" really means. – 2017-02-16

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As stated above there are two interpretations. In both case let's define $s = \sum_{i=1}^n k_i$.

  • Case 1, we assume that each person is unique namely that if Alice sits in seat 1 and Bob sits in seat 2 then this is a different combination than Alice in 2 and Bob in 1. Then the total possible combinations is indeed $\frac{s!}{(s-m)!}$.

  • Case 2, we assume that every person is the same namely that if Alice sits in seat 1 and Bob sits in seat 2 then this is the same as if Alice sits in 2 and Bob in 1. Then the total possible combinations is $\binom{s}{m}$.