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:

120

The correct answer is Option (4) → 120

The formula for combinations is:

$^nC_r = \frac{n!}{r!(n-r)!}$

Step 1: Select 3 men out of 6

We need to choose 3 men from the 6 available.

$^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \text{ ways}$

Step 2: Select 2 women out of 4

We need to choose 2 women from the 4 available.

$^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \text{ ways}$

Step 3: Find the Total Number of Ways

Since we need to perform both actions to form one committee, we multiply the results:

$\text{Total ways} = (^6C_3) \times (^4C_2)$

$\text{Total ways} = 20 \times 6 = 120$