The number of ways in which we can choose a committee from four men and six women, so that the committee includes atleast two men and exactly twice as many women as men is

Options

A. 94
B. 126
C. 128
D. None of these

Explanation

To solve the problem of choosing a committee from four men and six women, where the committee includes at least two men and exactly twice as many women as men, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 4 men and 6 women.
- The committee must include at least 2 men.
- The number of women must be exactly twice the number of men.

2. **Identifying Possible Cases**:
- Since the committee must include at least 2 men, we can have:
- Case 1: 2 men and 4 women
- Case 2: 3 men and 6 women
- We cannot have 4 men because that would require 8 women (which we do not have).

3. **Calculating for Case 1 (2 Men and 4 Women)**:
- The number of ways to choose 2 men from 4 is given by the combination formula \( \binom{n}{r} \):
\[ \text{Ways to choose 2 men} = \binom{4}{2} \]
- The number of ways to choose 4 women from 6 is:
\[ \text{Ways to choose 4 women} = \binom{6}{4} \]
- Therefore, the total ways for Case 1:
\[ \text{Total for Case 1} = \binom{4}{2} \times \binom{6}{4} \]

4. **Calculating for Case 2 (3 Men and 6 Women)**:
- The number of ways to choose 3 men from 4 is:
\[ \text{Ways to choose 3 men} = \binom{4}{3} \]
- The number of ways to choose 6 women from 6 is:
\[ \text{Ways to choose 6 women} = \binom{6}{6} \]
- Therefore, the total ways for Case 2:
\[ \text{Total for Case 2} = \binom{4}{3} \times \binom{6}{6} \]

5. **Calculating the Values**:
- For Case 1:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
\[ \binom{6}{4} = \binom{6}{2} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15 \]
\[ \text{Total for Case 1} = 6 \times 15 = 90 \]

- For Case 2:
\[ \binom{4}{3} = 4 \]
\[ \binom{6}{6} = 1 \]
\[ \text{Total for Case 2} = 4 \times 1 = 4 \]

6. **Final Calculation**:
- The total number of ways to form the committee is the sum of the totals from both cases:
\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} = 90 + 4 = 94 \]

### Conclusion:
The total number of ways to form the committee is **94**.