A Pair Apart

Imagine these $20$ students lined up in a row. Now the first person in the row can be paired up with any of the other $19$ students. That's $19$ ways. For each of these ways, look at the first person left in the row who has not yet been partnered up. That student has 17 ways to partner someone. So the first two pairings can be done in $19*17$ ways. For each of these ways, the first student in the row who doesn't have a partner can be paired up in 15 ways and so on. So thats $19*17*15*13*11*9*7*5*3*1=654729075$ ways.

Another way of solving. We can select a pair in $20C2$ ways. Then the next pair in $18C2$ ways, and the next in $16C2$ ways. and so on on till $2C2$ . That's $20C2*18C2*16C2*.......*2C2$ . However we are over-counting because we aren't interested in the order in which we picked the pairs. So we must divide by by the number of orderings in which you could have picked any set of ten pairs, and that's the number of permutations of ten objects, which is 10!