Counting permutations of students standing in line Say I have k students, four of them are Jack, Liz, Jenna and Tracy. I want to count the number of permutations in which Liz is standing left to Jack and Jenna is standing right to Tracy. I define $A = ${Liz is left to Jack} so $|A| = \frac{k!}{2}$. The same goes for $B$ with Jenna and Tracy. 
I know that $$|A \cap B| = |A| + |B| - |A \cup B|$$
But how do I find the union? I'm guessing it involves inclusion-exclusion, but I can't remember how exactly. 
Any ideas? Thanks!

There are $\dbinom{k}{2}$ ways to choose the positions that we will ultimately put Liz and Jack into. For each such choice, there is only one way to place Liz and Jack, since Liz is to the left of Jack. 
For choice we have made for Liz and Jack, there are $\dbinom{k-2}{2}$ ways to choose the positions occupied by the other two. And now there are $(k-4)!$ ways to choose the positions of the others, for a total of
$$\binom{k}{2}\binom{k-2}{2}(k-4)!.$$ 
