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.

We assume that a telephone conversations involve two people. So the question comes down to finding how to divide a group of $10$ into $5$ groups of $2$.
Line up the people in order of height, or student number.
Person $1$ can choose her partner in $9$ ways. For each such choice, the first person in the list who has not yet been partnered can choose her partner in $7$ ways. Now the first person not yet partnered can choose her partner in $5$ ways, and so on.
So the total number of ways is $(9)(7)(5)(3)(1)$. 
Another way: We choose $2$ people from the $10$, then choose $2$ people from the remaining $8$, and so on.
There are $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}$ ways to do this.  But the number just calculated grossly overcounts the number of ways to divide our people into telephone teams of $2$. This is because the product just obtained counts the each pairing $5!$ times, since it takes order of choosing into account. To get the right count, we must therefore divide by $5!%.
Remark: There are many other reasonable ways of doing the counting. In this problem, it is all too easy to overcount. 
