The correct answer is option (A):
455
Here's how to solve this problem, along with the reasoning behind it:
We need to form a committee of 5 members with at least one boy and at least one girl. A good approach is to calculate the total number of ways to form a committee *without* any restrictions, then subtract the cases where the committee fails to meet our criteria (i.e., has no boys or no girls).
1. **Total Unrestricted Committees:** We have 6 boys + 5 girls = 11 people in total. We need to choose a committee of 5. The number of ways to do this is given by the combination formula:
* ¹¹C₅ = (11!)/(5! * 6!) = (11 * 10 * 9 * 8 * 7) / (5 * 4 * 3 * 2 * 1) = 462
2. **Committees with No Boys (All Girls):** This is where the committee consists entirely of girls. We need to choose 5 girls from the 5 available.
* ⁵C₅ = (5!)/(5! * 0!) = 1 (There's only one way to choose all 5 girls)
3. **Committees with No Girls (All Boys):** This is where the committee consists entirely of boys. We need to choose 5 boys from the 6 available.
* ⁶C₅ = (6!)/(5! * 1!) = 6
4. **Subtract the Invalid Cases:** To get the number of committees with at least one boy and one girl, we subtract the cases with no boys and the cases with no girls from the total number of unrestricted committees:
* 462 (total committees) - 1 (all girls) - 6 (all boys) = 455
Therefore, there are 455 ways to form the committee with at least one boy and at least one girl.