combinations problem. If you have 5 women and 7 men, and you are after how many different groups of 2 women and 3 men you can form then i understand that the answer is:
${5\choose2} * {7\choose3}$ = 350 groups
but what if you are told that 2 men can't be grouped together then you have to look at the 35 possible groups of 3 from 7 and take away:
${2\choose2}*{5\choose1}$ = 5 groups 
then proceed as before. But where did this calculation come from i dont understand it?
Thanks for any help
 A: Let's say the men are $M_1,\cdots, M_7$ and that it's $M_1,M_2$ that can't be sat together.  
We already have the unrestricted total so we just want to subtract off the ones that have $M_1,M_2$ together. 
If a group contains $M_1,M_2$ then there are $5$ men remaining from which to choose the third man.  That the $\binom 51$ term.
There are no new rules on the women, so we still have $\binom 52$ ways to choose the women on the group.  
Thus there are $$\binom 51\times \binom 52$$ groups of $3$ men and $2$ women which contain both $M_1$ and $M_2$.  So the number of groups of $3$ men and $2$ women which do not contain both $M_1$ and $M_2$ is $$\binom 73\times \binom 52-\binom 51\times \binom 52=\binom 52\times \left(\binom 73-\binom 51\right)$$
Note: I'm not entirely sure how this connects to the answer you are reporting.
A: If the two distinguished men stand together, then there are obviously $5$ ways to choose the remaining man in the group.  The calculation $${2\choose2}{5\choose1}$$ is attempting to generalize this argument somehow, but I don't really think it succeeds.  Yes, if we were told that there were to be $4$ men in a group then the second factor would be ${5\choose2}$, so that's okay. 
It's hard for me to see what the first factor is supposed to generalize to. 
If we were told that there were $3$ men, no two of whom would stand together, then we could say that there are ${3\choose2}$ ways to choose two men would can't stand together, but we'd have to have to worry about the double-counting the cases where all three are in the final group.  
