Q.A committee of 5 members is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must contain exactly 3 men and 2 women?

The condition splits the selection into two independent combination choices — men and women — whose counts are then multiplied by the Fundamental Principle of Counting.

The committee must have exactly 3 men (from the 6 available) and exactly 2 women (from the 4 available), and since committee membership carries no internal ranking, both are combination choices.

Number of ways to choose 3 men from 6: $^{6}C_{3} = \dfrac{6!}{3!3!} = \dfrac{720}{6\times6}=20$.

Number of ways to choose 2 women from 4: $^{4}C_{2} = \dfrac{4!}{2!2!} = \dfrac{24}{2\times2}=6$.

Since choosing the men and choosing the women are independent tasks that both happen (not either-or), the Fundamental Principle of Counting says the two counts multiply:

$20 \times 6 = 120$ …

Adding the two combination counts ($20+6=26$) instead of multiplying them — since both a men-choice and a women-choice must happen together (not as alternatives), the …