6.2 Q-3

Exercise 6.2 — Question 3

Key Concept: Combinations

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

The number of ways to choose $r$ objects from $n$ distinct objects is:

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

Useful Properties

• $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$
• $\left(\genfrac{}{}{0ex}{}{n}{n}\right)=1$
• $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-r}\right)$ (symmetry property)
• $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ (Pascal's identity)

Worked Example (Typical Q3-style Problem)

Problem: In how many ways can a committee of 3 be selected from 8 people?

Solution:

Since order does not matter, we use combinations:

$\left(\genfrac{}{}{0ex}{}{8}{3}\right)=\frac{8!}{3!\cdot 5!}=\frac{8\times 7\times 6}{3\times 2\times 1}=\frac{336}{6}=56$

So there are 56 ways to form the committee.

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: Choosing players from a squad without regard to position.