6.2 Q-8

Exercise 6.2 — Question 8

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 $n$ distinct items (where order does not matter) is given by:

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

---

Solution

We need to choose 5 members from 8 people. Since the order of selection does not matter (a committee is a set, not a sequence), we use combinations:

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

Expanding:

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

$\boxed{56\text{ committees}}$

---

Why Combinations (Not Permutations)?

A committee of {Ali, Sara, Ahmed} is the same committee regardless of the order they were chosen — so we use $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$.

---

General Tip

For combination problems:

1. Identify $n$ (total items) and $r$ (items to choose).
2. Apply $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$.
3. Simplify by cancelling the larger factorial in the denominator.