Question

In how many ways a committee of 5 members can be selected from 6 men and 5 women, consisting 3 men and 2 women ?

Answer 1

Three men out of 6 men can be selected in $6_{C_3}$ ways. Two women out of 5 women can be selected in $5_{C_2}$ ways. Therefore, by the fundamental principle of counting, 3 men out of 6 men and 2 women out of 5 women can be selected in

$6_{C_3}\times 5_{C_2}=(\frac{6\times 5\times 4}{3\times 2\times 1}\times \frac{5\times 4}{2\times 1})=200ways$.