Number of ways to form commissions that must contain the same number of people from each group and males = females. We have a group A, with 12 members that 3 of them are female.
A group B, with 13 members that 5 of the are female.
What are the ways to form a commission of 8 people that contains:
4 people from group A and 4 people from group B and males are the same as females.
$\binom{12}{4} \times \binom{13}{4}=353925$, when people of gr. A = gr. B.
$\binom{8}{4} \times \binom{17}{4}=2380$, when male = female.
Till now I have find these but I don't know how to find the intersection.
 A: The possibilities to be added are:-A: $4$ males, $0$ females and B: $0$ males, $4$ females                                                     $\begin{pmatrix}9\\4\\\end{pmatrix}\begin{pmatrix}3\\0\\\end{pmatrix}\begin{pmatrix}9\\0\\\end{pmatrix}\begin{pmatrix}4\\4\\\end{pmatrix}=126$A: $3$ males, $1$ females and B: $1$ males, $3$ females                                                     $\begin{pmatrix}9\\3\\\end{pmatrix}\begin{pmatrix}3\\1\\\end{pmatrix}\begin{pmatrix}9\\1\\\end{pmatrix}\begin{pmatrix}4\\3\\\end{pmatrix}=9072$A: $2$ males, $2$ females and B: $2$ males, $2$ females                                                     $\begin{pmatrix}9\\2\\\end{pmatrix}\begin{pmatrix}3\\2\\\end{pmatrix}\begin{pmatrix}9\\2\\\end{pmatrix}\begin{pmatrix}4\\2\\\end{pmatrix}=23328$A: $1$ males, $3$ females and B: $3$ males, $1$ females                                                     $\begin{pmatrix}9\\1\\\end{pmatrix}\begin{pmatrix}3\\3\\\end{pmatrix}\begin{pmatrix}9\\3\\\end{pmatrix}\begin{pmatrix}4\\1\\\end{pmatrix}=3024$
Total $35550$.
