In how many ways can a comittee of 3 women and 4 men be chosen from 8 women and 7 men if two particular women refuse to serve on the committe together? Answer is 1750 but I have no clue how to get to it

Answer 1

Let's choose 3 women and 4 men from the 8 women and 7 men without any restrictions.
= C(8,3) x C(7,4) = ....

Now choose with the two particular women in place and th 4 men from the 7
That means we have to choose 1 more women from the remaining 6, the men situation does not change
= C(6,1) x C(7,4)

subtract this from the total found first, that should give you 1750
(worked for me)

Answer 2

men ... 7C4 possibilities ... 35

women ... 8C3 possibilities ... 56
... but 6 of the groups can't be used ... the incompatible pair with any of the other 6 women

35 * (56 - 6) = ?