6.1 Q-7

Exercise 6.1 — Question 7

Topic: Combinations

This question involves applying the combination formula to solve counting problems where order does not matter.

Key Formula

The number of ways to choose $r$ objects from $n$ distinct objects (without regard to order) is:

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

Properties Used

• $C(n,0)=1$
• $C(n,n)=1$
• $C(n,r)=C(n,n-r)$ (symmetry property)
• $C(n,r)+C(n,r-1)=C(n+1,r)$ (Pascal's identity)

Worked Example

Problem: In how many ways can a committee of 3 be selected from 7 people?

Solution:

Since order does not matter, use the combination formula:

$C(7,3)=\frac{7!}{3!(7-3)!}=\frac{7!}{3!\cdot 4!}$

$=\frac{7\times 6\times 5\times 4!}{3!\times 4!}=\frac{7\times 6\times 5}{3\times 2\times 1}=\frac{210}{6}=35$

There are 35 ways to form the committee.

Simplification Technique

When computing $C(n,r)$, cancel the larger factorial in the denominator with part of $n!$:

$C(n,r)=\frac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$

This avoids computing large factorials directly.