I've been working on a text on permutations and I came across this question that I have no idea how to even begin to solve.
There are 5 desks arranged one behind the other, and 10 students (let's call them A, B, C, ..., J). Each desk has 2 chairs so the students must be seated in pairs.
There are two restrictions on how they are positioned:
1. Student A must be seated in front of B.
2. Student B must be seated in front of C, D and E.
When I say in front, it doesn't have to be directly, just any desk in front.
How many possible seating arrangements are there?
You can sum over $B$’s desk. Let $k$ denote the number of $B$’s desk as counted from the front. Then $k\ge2$, since $A$ must sit in front of $B$, and $k\le3$, since there must be room for $C$ through $E$ behind $B$. Thus $k\in\{2,3\}$. There are $2$ spots for $B$, $2k-2$ spots for $A$, $10-2k$ spots for $C$ through $E$ and $5$ spots left for everyone else, so the total is
$$ \sum_{k=2}^32\cdot(2k-2)\cdot\frac{(10-2k)!}{(7-2k)!}\cdot5!=2\cdot5!\left(2\cdot\frac{6!}{3!}+4\cdot\frac{4!}{1!}\right)=80640\;. $$