A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Options

A. 10,10
B. 10,6
C. 6,6
D. 6,10

Explanation

To solve the problem step by step, we will break it down into two parts: first, calculating the total number of ways to form a committee of 3 persons from a group of 2 men and 3 women, and second, determining how many of those committees consist of 1 man and 2 women.

### Step 1: Calculate the total number of ways to form a committee of 3 persons.

1. **Identify the total number of people**:
- We have 2 men and 3 women.
- Total number of people = 2 + 3 = 5.

2. **Use the combination formula**:
- We need to choose 3 persons from these 5. The formula for combinations is given by:
\[ nCr = \frac{n!}{r!(n-r)!} \]
- Here, \( n = 5 \) and \( r = 3 \).
- So, we need to calculate \( 5C3 \).

3. **Calculate \( 5C3 \)**:
\[ 5C3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} \]
- Expanding the factorials:
\[ = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} \]
- The \( 3! \) cancels out:
\[ = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10 \]

### Total ways to form a committee of 3 persons: **10**

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### Step 2: Calculate the number of committees consisting of 1 man and 2 women.

1. **Choose 1 man from 2 men**:
- The number of ways to choose 1 man from 2 is given by:
\[ 2C1 = \frac{2!}{1!(2-1)!} = \frac{2!}{1! \cdot 1!} = 2 \]

2. **Choose 2 women from 3 women**:
- The number of ways to choose 2 women from 3 is given by:
\[ 3C2 = \frac{3!}{2!(3-2)!} = \frac{3!}{2! \cdot 1!} = 3 \]

3. **Multiply the combinations**:
- The total number of ways to form a committee with 1 man and 2 women is:
\[ \text{Total} = 2C1 \times 3C2 = 2 \times 3 = 6 \]

### Total committees consisting of 1 man and 2 women: **6**

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### Summary of Results:
- Total ways to form a committee of 3 persons: **10**
- Total committees consisting of 1 man and 2 women: **6**

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