Combinations trouble

Question

A class consists of 14 males and 12 females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of 6 men and 4 women be chosen from the class?

My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?

so (13,5) demonstrates that male.a in on the committee therefore (11,4) is female.a not being allowed on the committee. (13,5) male.a not allowed on committee and (11,4) is female.a on the committee.

Answer 1

My thought process so far is to find total combinations subtract the constraint scenarios.

Yes, but the constraint(or forbidden) scenario is to "select a commitee containing both A and B".   Count that and subtract it from the total.

You are allowed to select commitees containing only one from the twain (so don't subtract these).

Answer 2

You may construct the number of "forbidden" committees as follows:

• $A$ and $5$ other males and $B$ and 3 other females: $\color{blue}{\binom{13}{5}\cdot\binom{11}{3}}$

Subtracting from the total number of committees you get the "allowed" committees: $$\binom{14}{6}\cdot\binom{12}{4} - \color{blue}{\binom{13}{5}\cdot\binom{11}{3}}$$