In how many different ways can it be done ?
10 men and 8 women out of which 5men are teachers, 3 men doctors and 2 businessmen ,Among the women , 3 are teachers, 2 doctors , 2 researches and 1 social worker.
A committee of 4 in which at least 2 women are there

Options

A. 1260
B. 1820
C. 3060
D. 1890

Explanation

To solve the problem of forming a committee of 4 members with at least 2 women from a group of 10 men and 8 women, we can break it down into cases based on the number of women in the committee. The possible cases are:

1. **Case 1**: 2 women and 2 men
2. **Case 2**: 3 women and 1 man
3. **Case 3**: 4 women

Let's calculate the number of ways for each case.

### Case 1: 2 Women and 2 Men
- **Selecting 2 Women from 8**: The number of ways to choose 2 women from 8 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).

\[ C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]

- **Selecting 2 Men from 10**: The number of ways to choose 2 men from 10 is:

\[ C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]

- **Total Ways for Case 1**: Multiply the ways to choose women and men.

\[ \text{Total for Case 1} = C(8, 2) \times C(10, 2) = 28 \times 45 = 1260 \]

### Case 2: 3 Women and 1 Man
- **Selecting 3 Women from 8**:

\[ C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

- **Selecting 1 Man from 10**:

\[ C(10, 1) = 10 \]

- **Total Ways for Case 2**:

\[ \text{Total for Case 2} = C(8, 3) \times C(10, 1) = 56 \times 10 = 560 \]

### Case 3: 4 Women
- **Selecting 4 Women from 8**:

\[ C(8, 4) = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

- **Total Ways for Case 3**: Since there are no men in this case, the total is simply:

\[ \text{Total for Case 3} = C(8, 4) = 70 \]

### Final Calculation
Now, we add the total ways from all cases:

\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} + \text{Total for Case 3} \]

\[ \text{Total Ways} = 1260 + 560 + 70 = 1890 \]

Thus, the total number of different ways to form the committee is **1890**.