You’ve miscounted the ways to choose $3$ women willing to serve together.
There are $\binom83$ ways to choose $3$ women without any restriction. If we choose the two difficult women, there are $6$ ways to choose the third woman, so there are only $6$ unusable sets of $3$ women. Thus, the number of acceptable sets of $3$ women is $\binom83-6$, and the number of committees is
It’s possible to do it by counting the usable sets of $3$ women directly, but it’s a little harder. There are $\binom63$ ways to choose a group that has neither of the difficult women. To choose a group that has exactly one of them, we first choose which of the two we’ll include, and then we choose $2$ of the $6$ women who don’t care with whom they serve; this can be done in $2\cdot\binom62$ ways. We then get a total of
ways to form the committee.
To see exactly how you’ve miscounted, notice that your $\binom73$ corresponds to my $2\binom62$. You’ve picked one of the difficult women, set her aside, and chosen any $3$ from the remaining $7$. The problem here is two-fold. First, this count includes every group of $3$ women that does not include either of the difficult women, and you already counted those groups in your $\binom63$. Secondly, it doesn’t take into account the fact that there are $2$ ways to decide which difficult woman is not on the committee.