6.1 Q-5

Exercise 6.1 — Question 5: Combinations

Key Formula

The number of combinations of $n$ distinct objects taken $r$ at a time is:

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

where $0\le r\le n$.

Difference from Permutations: In combinations, order does not matter. The extra $r!$ in the denominator removes the effect of ordering:

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

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Useful Properties

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Worked Examples

Example 1: Evaluate ${}^8{C}_3$.

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

Example 2: In how many ways can a committee of 3 be chosen from 7 people?

Since order does not matter: ${}^7{C}_3=\frac{7!}{3!\cdot 4!}=\frac{7\times 6\times 5}{6}=35\text{ ways}$

Example 3: Simplify $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}$.

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}=\frac{(r-1)!(n-r+1)!}{r!(n-r)!}=\frac{n-r+1}{r}$

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Factorial Simplification (Supporting Skill)

Many combination problems require simplifying factorial expressions. Recall:

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

Example: $\frac{(n+1)!}{(n-1)!}=(n+1)\cdot n=n^2+n$