Out of 5 men and 2 women a committee of 3 persons is to be formed so as to include atleast one woman. The number of ways in which it can be done is
(i) 10
(ii) 25
(iii) 35
(iv) 45

Options

A. 10
B. 25
C. 35
D. 45

Answer

To solve the problem of forming a committee of 3 persons from 5 men and 2 women, ensuring that at least one woman is included, we can break it down into two cases:

### Step 1: Identify the Cases
We can have:
1. **Case 1:** 1 woman and 2 men
2. **Case 2:** 2 women and 1 man

### Step 2: Calculate Case 1 (1 Woman and 2 Men)
- **Choosing 1 woman from 2:** This can be done in \( \binom{2}{1} \) ways.
- **Choosing 2 men from 5:** This can be done in \( \binom{5}{2} \) ways.

Now, we calculate these values:
- \( \binom{2}{1} = 2 \)
- \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)

So, the total ways for Case 1 is:
\[ \text{Total for Case 1} = \binom{2}{1} \times \binom{5}{2} = 2 \times 10 = 20 \]

### Step 3: Calculate Case 2 (2 Women and 1 Man)
- **Choosing 2 women from 2:** This can be done in \( \binom{2}{2} \) ways.
- **Choosing 1 man from 5:** This can be done in \( \binom{5}{1} \) ways.

Now, we calculate these values:
- \( \binom{2}{2} = 1 \)
- \( \binom{5}{1} = 5 \)

So, the total ways for Case 2 is:
\[ \text{Total for Case 2} = \binom{2}{2} \times \binom{5}{1} = 1 \times 5 = 5 \]

### Step 4: Combine the Results
Now, we add the results from both cases to find the total number of ways to form the committee:
\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} = 20 + 5 = 25 \]

### Final Answer
The total number of ways to form a committee of 3 persons that includes at least one woman is **25**.