The I Eta Pi fraternity must choose a delegation of 3 senior members and 2 junior members for an annual interfraternity conference. If I Eta Pi has 6 senior members and 5 junior members, how many different delegations are possible?

Options

A. 100
B. 200
C. 150
D. 300

Answer

To solve the problem of choosing a delegation of 3 senior members and 2 junior members from the I Eta Pi fraternity, we can break it down into steps using the concept of combinations.

### Step-by-Step Solution:

1. **Identify the Total Members**:
- There are 6 senior members and 5 junior members in the fraternity.

2. **Determine the Number of Members to Choose**:
- We need to choose 3 senior members from the 6 available.
- We also need to choose 2 junior members from the 5 available.

3. **Use the Combination Formula**:
- The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
- Where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

4. **Calculate the Combinations for Senior Members**:
- For choosing 3 senior members from 6:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} \]
- Simplifying this:
\[ = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

5. **Calculate the Combinations for Junior Members**:
- For choosing 2 junior members from 5:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} \]
- Simplifying this:
\[ = \frac{5 \times 4}{2 \times 1} = 10 \]

6. **Calculate the Total Number of Different Delegations**:
- The total number of different delegations is the product of the combinations of senior and junior members:
\[ \text{Total Delegations} = \binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200 \]

### Final Answer:
The total number of different delegations possible is **200**.

---