Combinations - excluding one from each group I have 8 women and 6 men and need to form a committee with 3 women and 3 men.  If one man and one woman refuse to work together how many committees can be formed?

*

*Determine the number of committees excluding the 1 man and 1 woman: 
$\binom{7}{3}\binom{5}{3}  = 350$


*Determine the valid combinations including all the women and excluding the one man:
$\binom{8}{3} \binom{5}{2}  = 560$


*Determine the valid combinations including all the men and excluding the one woman:
$\binom{7}{2}\binom{6}{3}  = 441$


*Find the total: 
350 + 560 + 441 = 1351
Obviously this is wrong because the total number of committees, if everyone is willing to work with everyone is:
$\binom{8}{3}\binom{6}{3}  = 1120$
Can anyone help with how I am supposed to be solving this?
Thanks
 A: If you want to do it your way, the correct count would be:

*

*The committees that contain neither the woman nor the man. $\binom{6-1}{3}\binom{8-1}{3}$

*The committees that contain the woman but not the man. $\binom{6-1}{3}\binom{8-1}2.$

*The committees that contain the man but not the woman. $\binom{6-1}2\binom{8-1}3.$
Note the top parts of the binomials are always $6-1$ and $8-1,$ never $6$ or $8.$
Of course, if we add:


*The committees that contain both the man and the woman $\binom{6-1}2\binom{8-1}2.$
We get all possible committees, $\binom63\binom83.$ So the easier way is to subtract 4. from the total sum.
A: Your issue is that you've counted many of the possible committees multiple times. For instance, a committee which excludes both of the given people will be counted in each of the three expressions you give.
An easier approach is to count the total number of committees with no restriction, then subtract the committees which contain both of those people. That is, (8C3)(6C3)-(7C2)(5C2).
A: Let $1,2,3,4,5,6,7,8$ be the women, and the men are denoted similarly, but with a hat, $\hat1$, $\hat2$, $\hat3$, $\hat4$, $\hat5$, $\hat6$.
The $1$ is not willing to work with $\hat 1$.
Now let us look at the counting strategy.
Where do we count the configuration $2,3,4; \hat2,\hat3,\hat 4$?

*

*Well, at point one at any rate, it is one valid case among the many
$\binom 73\binom 53$, since $1$ and $\hat 1$ are both excluded.

*Well, at point two also, it is one valid case among the many
$\binom 83\binom 52$ cases, since $\hat 1$ is excluded.

*Well, at point three again, it is one valid case among the many
$\binom 72\binom 63$ cases, since $1$ is excluded.


For a correct answer, count all possible commitees, there are $\binom 83\binom 63=1120$ of them, and subtract those where the pair $1,\hat 1$ is in the commitee, there are $\binom 72\binom 52=210$ of them. So the answer is
$$
\binom 83\binom 63
-
\binom 72\binom 52
=1120
-210
=910\ .
$$
