Question

How many ways can a committee chairman choose 4 people for a subcommittee if there are 12 members on the whole committee?

Answer 1

A group of k elements can be chosen from a group of n elements in $\frac{n!}{k!(n-k)!}$ ways.

$(^{12} _4)=\frac{12!}{4!(12-4)!}=\frac{12!}{4! \times 8!}=\frac{8! \times 9 \times 10 \times 11 \times 12}{2 \times 3 \times 4 \times 8!}=\frac{9 \times 10 \times 11 \times 12}{2 \times 3 \times 4}= \\ \\ =3 \times 5 \times 11 \times 3=495$

The answer is B. 495.