Form a group of 6 men and 4 women a committee of 4 persons is to be formed:
In how many different ways can it be done so that the committee has at least 2 men?

Options

A. 210
B. 225
C. 195
D. 185

Answer

To solve the problem of forming a committee of 4 persons from a group of 6 men and 4 women, ensuring that the committee has at least 2 men, we can break it down into three cases:

### Step-by-Step Solution:

1. **Identify the Cases**: We need to consider the following cases for the committee composition:
- Case 1: 2 men and 2 women
- Case 2: 3 men and 1 woman
- Case 3: 4 men and 0 women

2. **Calculate for Case 1 (2 Men and 2 Women)**:
- The number of ways to choose 2 men from 6 is given by the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).
- Thus, the number of ways to choose 2 men from 6 is:
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
- The number of ways to choose 2 women from 4 is:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
- Therefore, the total for Case 1 is:
\[ 15 \times 6 = 90 \]

3. **Calculate for Case 2 (3 Men and 1 Woman)**:
- The number of ways to choose 3 men from 6 is:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
- The number of ways to choose 1 woman from 4 is:
\[ \binom{4}{1} = 4 \]
- Therefore, the total for Case 2 is:
\[ 20 \times 4 = 80 \]

4. **Calculate for Case 3 (4 Men and 0 Women)**:
- The number of ways to choose 4 men from 6 is:
\[ \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
- There are no women chosen, so the total for Case 3 is:
\[ 15 \times 1 = 15 \]

5. **Add All Cases Together**:
- Now, we sum the totals from all three cases:
\[ 90 + 80 + 15 = 185 \]

### Final Answer:
The total number of different ways to form a committee of 4 persons with at least 2 men is **185**.