How many ways are there for 10 people to have five simultaneous telephone conversations? So what I'm thinking is 
$$
\binom{10}{2}\binom{8}{2} \ldots \binom{2}{2}
$$
But that's far from what the answer is.

Presumably this is to be interpreted as asking for the number of different ways to pair up $10$ people. Imagine numbering them $1$ through $10$. There are $9$ choices for the person to be paired with Nr. $1$. Once that pair has been determined, find a partner for the lowest-numbered person remaining; that will be either Nr. $2$ or Nr. $3$, depending on who Nr. $1$’s partner is. Either way, there are $7$ possible choices for his partner, because $3$ people are out of the running: he himself, and the first pair. That gives us $9\cdot7$ ways to choose the first two pairs. Now find a partner for the lowest-numbered person remaining: there are $5$ available choices, since we have to exclude the first two pairs and the person for whom we’re finding a partner. Putting the pieces together, we can form the first three pairs in $9\cdot7\cdot5$ ways.
Can you finish the analysis from there?
Added: Judging by your calculation, you’re thinking of choosing a pair, then a pair from the remaining $8$, and so on. The problem is that you might choose the same set of $5$ pairs in $5!$ different orders, so you’re overcounting by a factor of $5!$. And indeed,
$$\binom{10}2\binom82\dots\binom22=\frac{10!}{2^5}=5\cdot9\cdot4\cdot7\cdot3\cdot5\cdot2\cdot3\cdot1\cdot1=5!(9\cdot7\cdot5\cdot3\cdot1)\;.$$
