A committee of five members is to formed out of 3 trainees, 4 professors and 6 research associates .In how many different ways can this be done if :
The committee should have all 4 professors and 1 research associate or all 3 trainess and professors?

Options

A. 12
B. 13
C. 24
D. 52

Explanation

To solve the problem of forming a committee of five members from 3 trainees, 4 professors, and 6 research associates under the given conditions, we will break it down into two cases:

### Case 1: The committee has all 4 professors and 1 research associate.

1. **Select Professors**: Since we need all 4 professors, there is only 1 way to choose them, which is \( \binom{4}{4} = 1 \).

2. **Select Research Associate**: We need to select 1 research associate from the 6 available. The number of ways to do this is \( \binom{6}{1} = 6 \).

3. **Total Ways for Case 1**: Multiply the number of ways to select professors and research associates:
\[ \text{Total Ways for Case 1} = \binom{4}{4} \times \binom{6}{1} = 1 \times 6 = 6. \]

### Case 2: The committee has all 3 trainees and 2 professors.

1. **Select Trainees**: Since we need all 3 trainees, there is only 1 way to choose them, which is \( \binom{3}{3} = 1 \).

2. **Select Professors**: We need to select 2 professors from the 4 available. The number of ways to do this is \( \binom{4}{2} \). Calculating this:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6. \]

3. **Total Ways for Case 2**: Multiply the number of ways to select trainees and professors:
\[ \text{Total Ways for Case 2} = \binom{3}{3} \times \binom{4}{2} = 1 \times 6 = 6. \]

### Final Calculation: Total Number of Ways

Now, we add the total ways from both cases:
\[ \text{Total Ways} = \text{Total Ways for Case 1} + \text{Total Ways for Case 2} = 6 + 6 = 12. \]

### Conclusion

The total number of different ways to form the committee under the given conditions is **12**.

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