Exercise 6.1 — Question 2

Topic: Factorial Notation and Simplification

This exercise focuses on simplifying expressions involving factorials and converting products of consecutive integers into factorial form.

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Key Property: Factorial Recursion

The fundamental recursive property of factorials is:

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

Applying this repeatedly allows cancellation in fraction expressions:

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

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Worked Examples

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

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

Example 2: Simplify $\frac{n!}{(n-3)!}$

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

Example 3: Write $n(n-1)(n-2)$ as a factorial expression

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

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

Example 4: Simplify $\frac{(2n)!}{(2n-2)!}$

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

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General Rule

For any positive integer $k$:

$\frac{n!}{(n-k)!}=n(n-1)(n-2)\cdots (n-k+1)(k\text{ factors})$

This is also the formula for the number of permutations: $P(n,k)=\frac{n!}{(n-k)!}$