Question

In how many ways can a committee of 5 people be chosen out of 9 people?

a. 254
b. 126
c. 120
d. 131

Answer 1

In how many ways can a committee of 5 people be chosen out of 9 people?

a. 254 b. 126 c. 120 d. 131

Solution

The number of k-combinations from a given set $S$ of $n$ elements is often denoted in elementary combinatorics texts by $C(n, k)$, where

$$C(n, k) = \frac{n!}{k! (n - k)!} = \frac{n(n - 1)(n - 2) \dots (n - k + 1)}{k!}.$$

$n!$ (n-factorial) is the product of all positive integers less than or equal to $n$ ($n! = 1 * 2 * \ldots * n$).

The number of possible outcomes of selecting 5 people from 9 is

$$C(9,5) = \frac{9 * 8 * 7 * 6}{1 * 2 * 3 * 4} = 126$$

Answer: b. 126.