Exercise 6.1 — Fundamental Counting Principle & Factorials

Key Concept: Factorials

The factorial of a non-negative integer $n$ is defined as:

$n!=n\times (n-1)\times (n-2)\times \cdots \times 2\times 1$

with the special case $0!=1$.

Recursive property: $n!=n\times (n-1)!$

This property is essential for simplifying factorial expressions.

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Simplifying Factorial Expressions

Type 1: Simplify $\frac{n!}{(n-r)!}$

Example: Simplify $\frac{n!}{(n-2)!}$

$\frac{n!}{(n-2)!}=\frac{n\cdot (n-1)\cdot (n-2)!}{(n-2)!}=n(n-1)$

Example: Simplify $\frac{(n+1)!}{(n-1)!}$

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

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Type 2: Express a product as a factorial ratio

Example: Write $n(n-1)(n-2)$ as a factorial expression.

Multiply and divide by $(n-3)!$:

$n(n-1)(n-2)=\frac{n(n-1)(n-2)\cdot (n-3)!}{(n-3)!}=\frac{n!}{(n-3)!}$

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Fundamental Counting Principle

If one event can occur in $m$ ways and a second independent event can occur in $n$ ways, then the two events together can occur in $m\times n$ ways.

Example: If you have 3 shirts and 4 trousers, the number of different outfits = $3\times 4=12$.

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Worked Examples

Q1(i): Simplify $\frac{n!}{(n-3)!}$

$\frac{n!}{(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}=n(n-1)(n-2)$

Q1(ii): Simplify $\frac{(2n)!}{(2n-2)!}$

$\frac{(2n)!}{(2n-2)!}=\frac{(2n)(2n-1)(2n-2)!}{(2n-2)!}=(2n)(2n-1)=4n^2-2n$

Q1(iii): Write $\frac{1}{6!}+\frac{1}{7!}$ as a single fraction

$\frac{1}{6!}+\frac{1}{7!}=\frac{7}{7!}+\frac{1}{7!}=\frac{8}{7!}$

(Since $7\times 6!=7!$, so $\frac{1}{6!}=\frac{7}{7!}$)

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