Exercise 6.2 — Question 23

Problem

How many committees of 5 members can be formed from a group of 8 people?

Key Formula

The number of ways to choose $r$ items from $n$ items (without regard to order) is given by:

$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$

Solution

We need to choose 5 members from 8 people. Order does not matter (a committee is a combination, not a permutation).

$\left(\genfrac{}{}{0ex}{}{8}{5}\right)=\frac{8!}{5!(8-5)!}=\frac{8!}{5!\cdot 3!}$

Expanding:

$=\frac{8\times 7\times 6\times 5!}{5!\times 3\times 2\times 1}=\frac{8\times 7\times 6}{6}=\frac{336}{6}=56$

Answer: 56 committees can be formed.

Why Combination (Not Permutation)?

• A permutation is used when order matters (e.g., arranging people in seats).
• A combination is used when order does not matter (e.g., selecting a committee, a team, or a group).
• Since committee members have no ranked positions, we use $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$.

General Approach for Combination Problems

1. Identify $n$ (total items) and $r$ (items to choose).
2. Confirm order does not matter → use combination formula.
3. Apply $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$.
4. Simplify by cancelling the larger factorial.