In how many ways can a committee of 5 made out 6 men and 4 women containing atleast one woman?

Options

A. 246
B. 222
C. 18
D. None of these

Answer

To solve the problem of forming a committee of 5 members from 6 men and 4 women, ensuring that there is at least one woman in the committee, we can follow these steps:

### Step 1: Calculate the Total Ways to Form a Committee of 5
First, we calculate the total number of ways to select a committee of 5 from the total of 10 people (6 men + 4 women) without any restrictions. This can be calculated using the combination formula:

\[ \text{Total ways} = \binom{10}{5} \]

### Step 2: Calculate the Ways to Form a Committee with No Women
Next, we need to find the number of ways to form a committee of 5 members that contains no women (i.e., all men). This can be calculated as:

\[ \text{Ways with no women} = \binom{6}{5} \]

### Step 3: Calculate the Ways with At Least One Woman
To find the number of ways to form a committee with at least one woman, we can subtract the number of ways to form a committee with no women from the total ways calculated in Step 1:

\[ \text{Ways with at least one woman} = \text{Total ways} - \text{Ways with no women} \]

### Step 4: Calculate Each Component
Now, we calculate each of these components:

1. **Total ways to select 5 from 10**:
\[ \binom{10}{5} = \frac{10!}{5! \cdot (10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

2. **Ways to select 5 from 6 men**:
\[ \binom{6}{5} = \frac{6!}{5! \cdot (6-5)!} = 6 \]

### Step 5: Final Calculation
Now we substitute the values back into the equation from Step 3:

\[ \text{Ways with at least one woman} = 252 - 6 = 246 \]

### Conclusion
Thus, the total number of ways to form a committee of 5 members containing at least one woman is **246**.