How many ways are there to 10 people sit in a row of 10 chairs if exactly 5 of them are sited on chairs with a number on ticket? I thought that the result is ${10 \choose 5} \cdot 5!$ (first we select people that are sited properly and the rest are sited arbitrary on the free places). But an official solution is 11088. Any ideas? Ok, I figured out. I'll leave the problem on site anyway.

There are $\binom{10}{5}$ ways to choose those that sit in their assigned chairs and $\mathcal{D}(5)=44$ ways to derange the other $5$, so you get $\binom{10}{5}\mathcal{D}(5)=11088$, where $\mathcal{D}(n)$ is the number of Derangements of a set of $n$ elements.