6.2 Q-10

Exercise 6.2 — Question 10

Problem

This question involves selecting groups (combinations) where order does not matter. Use the combination formula:

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

Key Concept: Combinations

A combination is a selection of $r$ objects from $n$ distinct objects where order does not matter.

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

Important identities:

• ${}^n{C}_0=1$
• ${}^n{C}_n=1$
• ${}^n{C}_r{=}^n{C}_{n-r}$ (symmetry property)
• ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$ (Pascal's identity)

Worked Example (Typical Q10 Type)

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

Solution:

Since order does not matter (a committee is a combination):

${}^8{C}_3=\frac{8!}{3!(8-3)!}=\frac{8!}{3!\cdot 5!}=\frac{8\times 7\times 6}{3\times 2\times 1}=\frac{336}{6}=56$

There are 56 ways to form the committee.

Real-World Applications