I failed to mention this in my initial post. I made a structural assumption in that the elements in groups A and B can be distinguished in some way. For example, if there are 2 groups of male and female that we can call A and B, we can assign 4 separate males into group A and 2 separate females into group B. If that's not the case, then what I've mentioned above is incorrect. With this assumption, your elements in groups A and B can essentially be numbered:

A_1, A_2, A_3, ..., A_10

B_1, B_2, B_3, ... B_16

In the reduced example I provided, there are 2 elements in A and 3 in B:

A_1, A_2

B_1, B_2, B_3

C(5, 3) states there are 10 possible combinations:

A_1, A_2, B_1

A_1, A_2, B_2

A_1, A_2, B_3

A_1, B_1, B_2

A_1, B_1, B_3

A_1, B_2, B_3

A_2, B_1, B_2

A_2, B_1, B_3

A_2, B_2, B_3

B_1, B_2, B_3

Of these 10, only 6 match the pattern of 1 A and 2 B's. We can then conclude that the probability of getting 1 A and 2 B's is 6/10 = 0.6. You should be able to generalize this to your problem.