Exercise 6.2 — Question 5

Problem

In how many ways can a committee of 3 members be selected from a group of 8 people?

Solution

We need to choose 3 people from 8, where order does not matter — this is a combination problem.

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

Substituting $n=8$ and $r=3$:

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

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

Answer: There are 56 ways to select a committee of 3 from 8 people.

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Key Concept: Combinations

A combination is a selection of objects where order does not matter.

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

where:

• $n$ = total number of objects
• $r$ = number of objects to be chosen
• $n\ge r\ge 0$

Important Properties

Difference: Permutation vs Combination

• Permutation ${}^n{P}_r$: order matters (arrangements)
• Combination ${}^n{C}_r$: order does not matter (selections)

${}^n{P}_r=\frac{n!}{(n-r)!}{}^n{C}_r=\frac{n!}{r!(n-r)!}$

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Real-World Application

Combinations are used in:

• Lottery odds: Choosing 6 numbers from 49 gives $\left(\genfrac{}{}{0ex}{}{49}{6}\right)=13,983,816$ possible tickets.
• DNA sequences: Selecting bases to form codons.
• Team selection: Picking players for a sports team from a squad.