A committee of 6 members is to be selected from a group of 8 men and 6 women in such a way that at least 3 men are there in the committee. In how many different ways can it be done ?

Options

A. 2506
B. 2534
C. 1120
D. 1050

Answer

To solve the problem of selecting a committee of 6 members from a group of 8 men and 6 women with the condition that there are at least 3 men in the committee, we can break down the solution into different cases based on the number of men selected.

### Step 1: Define the Cases
We can have the following cases based on the number of men in the committee:
1. 3 men and 3 women
2. 4 men and 2 women
3. 5 men and 1 woman
4. 6 men and 0 women

### Step 2: Calculate the Combinations for Each Case

**Case 1: 3 Men and 3 Women**
- Number of ways to choose 3 men from 8: \( \binom{8}{3} \)
- Number of ways to choose 3 women from 6: \( \binom{6}{3} \)
- Total for this case:
\[ \binom{8}{3} \times \binom{6}{3} \]

**Case 2: 4 Men and 2 Women**
- Number of ways to choose 4 men from 8: \( \binom{8}{4} \)
- Number of ways to choose 2 women from 6: \( \binom{6}{2} \)
- Total for this case:
\[ \binom{8}{4} \times \binom{6}{2} \]

**Case 3: 5 Men and 1 Woman**
- Number of ways to choose 5 men from 8: \( \binom{8}{5} \)
- Number of ways to choose 1 woman from 6: \( \binom{6}{1} \)
- Total for this case:
\[ \binom{8}{5} \times \binom{6}{1} \]

**Case 4: 6 Men and 0 Women**
- Number of ways to choose 6 men from 8: \( \binom{8}{6} \)
- Number of ways to choose 0 women from 6: \( \binom{6}{0} \)
- Total for this case:
\[ \binom{8}{6} \times \binom{6}{0} \]

### Step 3: Calculate the Values
Now we will compute the values for each case.

1. **Case 1:**
\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Total for Case 1: \( 56 \times 20 = 1120 \)

2. **Case 2:**
\[ \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]
\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]
Total for Case 2: \( 70 \times 15 = 1050 \)

3. **Case 3:**
\[ \binom{8}{5} = \binom{8}{3} = 56 \]
\[ \binom{6}{1} = 6 \]
Total for Case 3: \( 56 \times 6 = 336 \)

4. **Case 4:**
\[ \binom{8}{6} = \binom{8}{2} = 28 \]
\[ \binom{6}{0} = 1 \]
Total for Case 4: \( 28 \times 1 = 28 \)

### Step 4: Add All Cases Together
Now, we sum the totals from all cases:
\[ 1120 + 1050 + 336 + 28 = 2534 \]

### Final Answer
The total number of different ways to form the committee is **2534**.

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