Given:
A committee has seven men and four women.
four people are selected to go to a conference
We will use the combinations as follows
the number of ways to choose the four people are:
[tex]7C4+7C3+7C2+7C1+7C0[/tex]The rule of combinations is:
[tex]\text{nCr}=\frac{n!}{(n-r)!\cdot r!}[/tex]so, the number of ways will be:
[tex]35+35+21+7+1=99[/tex]The number of ways to make a group of two men and two women will be:
[tex]7C2=21[/tex]So, the chance that the group is two men and two women =
[tex]\frac{21}{99}=\frac{7}{33}[/tex]so, the answer will be 7/33
