If a group has 14 men and 11 women, how many different teams can be made with $6$ people that contains exactly $4$ women? 
A group has $14$ men and $11$ women.
  (a) How many different teams can be made with $7$ people?
  (b) How many different teams can be made with $6$ people that contains exactly $4$ women?

Answer key to a is $257$ but I can't figure out how to get $257$? There's no answer key to b though, but here's my attempt:
$$\binom{25}{6} - \left[\binom{14}{6} + \binom{14}{5}  + \binom{14}{4} + \binom{14}{3} + \binom{14}{1} + \binom{11}{6} + \binom{11}{5}\right]$$
What I'm trying to do here is subtracting all men, all women, 5 men, 4 men, 3 men, 1 men, and 5 women team from all possible combination of team. 
Thanks

Hint:
For the first one, number of ways you can choose $k$ woman and $m$ men= $11\choose k$+ $14\choose m$, such that $k+m=7$.
For the second one, number ways you can choose $4$ women= $14\choose4$
AND number of choosing $2$ men= $11 \choose 2$, when you have an AND, you gotta multiply them.
