A committee of 5 members is to be selected from among 6 boys and 5 girls. Determine the number of ways of selecting the committee if it is to consist of at least one boy and one girl.

Answer

To solve the problem of selecting a committee of 5 members from 6 boys and 5 girls with the condition that there is at least one boy and one girl, we can break it down into several steps:

### Step-by-Step Solution:

1. **Total Members Available**:
We have 6 boys and 5 girls, making a total of 11 members (6 boys + 5 girls).

2. **Total Ways to Select 5 Members**:
First, we calculate the total number of ways to select any 5 members from the 11 available members without any restrictions. This can be calculated using the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

\[ \text{Total ways} = \binom{11}{5} \]

3. **Calculate \( \binom{11}{5} \)**:
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!} \]
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

4. **Exclusion of Invalid Cases**:
We need to exclude the cases where there are no boys or no girls in the committee.

- **Case 1: All Boys**: Selecting 5 boys from 6.
\[ \text{Ways to select 5 boys} = \binom{6}{5} = 6 \]

- **Case 2: All Girls**: Selecting 5 girls from 5.
\[ \text{Ways to select 5 girls} = \binom{5}{5} = 1 \]

5. **Total Invalid Cases**:
The total number of invalid cases (all boys or all girls) is:
\[ \text{Invalid cases} = 6 + 1 = 7 \]

6. **Valid Cases**:
Now, we subtract the invalid cases from the total cases to find the valid cases.
\[ \text{Valid cases} = \text{Total ways} - \text{Invalid cases} = 462 - 7 = 455 \]

### Final Answer:
The total number of ways to select the committee consisting of at least one boy and one girl is **455**.

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