# Counting ways of sitting in adjacent seats

In how many ways can $m$ people entering a theatre be seated in two rows, each containing $n$ seats with the condition that no two sit in adjacent seats in the first row?

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First calculate how many ways $j$ seats can be occupied from $n$: this is ${n \choose j}$ if $n\ge j$.

Then calculate how many ways $k$ seats can be occupied from $n$ with none adjacent: this is ${n+1-k \choose k}$ if $n+1 \ge 2k$

So the answer to the original question is $$\sum_{k= \max(0,m-n)}^{\min \left(m ,\lfloor(n+1)/2\rfloor \right)} {n+1-k \choose k}{n \choose m-k}$$ if $2m \ge 3n+1$, and $0$ otherwise.

Multiply by $m!$ if order of the individuals matters.

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Then figure out, given any decision about which $m$ seats to use, how many ways there are to distribute the $m$ people among them. That is a standard problem.