A committee of 3 experts is to be selected out of a panel of 7 persons, three of them are engineers, three of them are managers and one is both engineer and manager. In how many ways can the committee be selected if it must have at least an engineer and a manager ?

Options

A. 33
B. 22
C. 11
D. 66

Answer

To solve the problem of selecting a committee of 3 experts from a panel of 7 persons (3 engineers, 3 managers, and 1 person who is both an engineer and a manager), while ensuring that the committee has at least one engineer and one manager, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Groups**:
- We have 3 engineers (E1, E2, E3).
- We have 3 managers (M1, M2, M3).
- We have 1 person who is both an engineer and a manager (let's call this person EM).

2. **Total Combinations Without Restriction**:
- The total ways to select any 3 members from 7 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Therefore, \( C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \).

3. **Calculate Invalid Combinations**:
- We need to exclude the combinations that do not meet the requirement of having at least one engineer and one manager.
- The invalid combinations are:
- All engineers: Selecting 3 engineers from 3 (which is not allowed).
- \( C(3, 3) = 1 \) (only one way to select all engineers).
- All managers: Selecting 3 managers from 3 (which is also not allowed).
- \( C(3, 3) = 1 \) (only one way to select all managers).

4. **Calculate Valid Combinations**:
- Total invalid combinations = 1 (all engineers) + 1 (all managers) = 2.
- Therefore, the valid combinations = Total combinations - Invalid combinations = \( 35 - 2 = 33 \).

5. **Final Count**:
- The total number of ways to select a committee of 3 experts with at least one engineer and one manager is **33**.

### Conclusion:
The total number of ways to select the committee is **33**.