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Solve the problem.

There are 12 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?

To determine the number of different subcommittees of 3 members that can be formed from a board of 12 members, we use the combination formula:

\[
C(n, k) = \frac{n!}{k!(n-k)!}
\]

where \( n \) is the total number of items, and \( k \) is the number of items to choose.

In this problem, \( n = 12 \) and \( k = 3 \).

\[
C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12!}{3! \cdot 9!}
\]

First, calculate the factorials involved:

\[
12! = 12 \times 11 \times 10 \times 9!
\]

Since \( 9! \) cancels out, we get:

\[
\frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220
\]

Therefore, the number of different subcommittees possible is \( \boxed{220} \).

Would you like more details or have any questions about this solution?

Here are some related questions to consider:
1. What is the combination formula and how is it derived?
2. How would the solution change if the subcommittee size was different?
3. What are permutations and how do they differ from combinations?
4. Can you calculate the number of subcommittees if there were 15 members instead of 12?
5. How does the combination formula apply to real-world situations outside of committees?
6. What would be the impact on the result if the selection was without replacement?
7. How can we verify the solution using a different method?
8. What are factorials, and how are they calculated?

**Tip:** In combinatorics, always ensure you understand whether the problem requires combinations (order doesn't matter) or permutations (order matters) to apply the correct formula.

Learn how to calculate the number of different subcommittees possible from a board of 12 members. Using the combination formula, C(12, 3), find out the total combinations. Suitable for students in Grades 9-12 studying combinatorics and permutations.

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