Out of 6 men and 4 women, a committee of 5 members is to be formed so that it has 2 women and 3 men. In how many different ways can it be done:

The committee needs 5 members with exactly 2 women and 3 men from a group of 6 men and 4 women.

Since the order of selection doesn't matter for a committee, combinations are used.

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The number of ways to choose 2 women from 4 women:

$$C(4,2) = \dfrac{4!}{2! \times 2!}$$C(4,2)=2!×2!4!​

$$C(4,2) = \dfrac{4 \times 3}{2 \times 1}$$C(4,2)=2×14×3​

$$C(4,2) = \dfrac{12}{2}$$C(4,2)=212​

$$C(4,2) = 6$$C(4,2)=6

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The number of ways to choose 3 men from 6 men:

$$C(6,3) = \dfrac{6!}{3! \times 3!}$$C(6,3)=3!×3!6!​

$$C(6,3) = \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1}$$C(6,3)=3×2×16×5×4​

$$C(6,3) = \dfrac{120}{6}$$C(6,3)=6120​

$$C(6,3) = 20$$C(6,3)=20

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The total number of ways to form the committee:

Total ways $= C(4,2) \times C(6,3)$=C(4,2)×C(6,3)

Total ways $= 6 \times 20$=6×20

Total ways $= 120$=120

Therefore, the committee can be formed in 120 different ways.