A game of tennis, in which one boy and one girl are matched against a similar pair, is to be made up from \(5\) boys and \(3\) girls. In how many ways can this be done?

There are five ways to choose the first boy.

Once he’s chosen, there are four possible boys left to play against him.

We have three possibilities for the first girl, and two for the second girl.

Thus we appear to have \(5 \times 4 \times 3 \times 2 = 120\) possible matches.

But we must note that the match in which boy \(A\) and girl \(A\) play against boy \(B\) and girl \(B\) is the same as the match where boy \(B\) and girl \(B\) play against boy \(A\) and girl \(A\).

So we counted each match twice above, and therefore we have \(\dfrac{120}{2} = 60\) different ways to organise these matches.

In how many of the games will a specified boy be playing?

If boy \(A\) is the specified one, then there are four boys left who can play against him.

As before, there are three possibilities for the first girl, and two for the second one.

Note that in this case we don’t count any matches twice, since we’ve chosen the specified boy first. So there are \(4 \times 3 \times 2 = 24\) such matches.

In how many of the games will he be opposing a second specified boy?

Let’s say boy \(A\) is the first boy, and \(B\) is the second boy.

There are now three possibilities for the first girl, and two for the second. Again, we do not count any match twice here, so there are \(3 \times 2 = 6\) such matches.