Ways to choose players in tennis match

Question

In how many ways can a game of tennis doubles be played from $3$ men and $4$ women when each team contains $1$ man and $1$ woman?

Answer 1

It would simply be $\binom31$ ways to choose $1$ man (for the first team) and $\binom41$ ways to choose $1$ woman from $4$ women. Thus you get $\binom31 \cdot \binom41$ = $3 \cdot 4$ = $12$ ways to choose the first team.

The second team can be made in $\binom21$ $\Bigl($ ways to choose men $\Bigl)$ and $\binom31$ ways to choose ways, women. Thus you get, $12 \cdot 6$ = $72$ ways of choosing teams.

Now divide the answer by $2$ because this includes both sides being different matches. Thus we get $36$ ways.

Answer 2

Hint: (assuming that which side of the court each time starts on is irrelevant)

Choose the two men and the two women simultaneously.

Then, choose which of the women is partnered with the younger of the two men.

Apply rule of product and arrive at an answer.