Question

Two committee members are chosen for a planning board out of twelve applicants. How many combinations are possible?

Answer 1

To determine the number of ways to choose 2 committee members from a pool of 12 applicants, we can use the formula for combinations, which is given by:

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

where:
- \( n \) is the total number of items (in this case, applicants),
- \( r \) is the number of items to choose (in this case, committee members), and
- \( ! \) denotes factorial, the product of all positive integers up to that number.

In this scenario, we have \( n = 12 \) and \( r = 2 \). Plugging these values into the combination formula:

\[
C(12, 2) = \frac{12!}{2!(12 - 2)!} = \frac{12!}{2! \cdot 10!}
\]

Calculating the factorials involved:

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

Thus, we can simplify:

\[
C(12, 2) = \frac{12 \times 11 \times 10!}{2! \times 10!}
\]

The \( 10! \) in the numerator and denominator cancels out:

\[
C(12, 2) = \frac{12 \times 11}{2!}
\]

Now, we compute \( 2! \):

\[
2! = 2 \times 1 = 2
\]

Plugging this back into the equation:

\[
C(12, 2) = \frac{12 \times 11}{2} = \frac{132}{2} = 66
\]

Thus, the number of combinations possible to choose 2 committee members out of 12 applicants is

\[
\boxed{66}
\]