This is a two stage combination problem (we know its combination because the order doesn't matter).

First we want to find out, out of 4 candidates, how many combinations of 2 men there are:

Ans = N!/(k!(N-k)!) = 4!/(2!(4-2)!) = 6 different men combinations

Second, we want to calculate, out of 3 candidates, how many women combinations there are:

(same formula) = 3!/(2!(3-2)1) = 3

Since each man combo can have any of the 3 woman combination, we multiply to get 18 total different combinations.

Hope this helps, good luck.