### Factorials

#### Definition

A factorial of a positive number $n$ is defined to be the product of all positive whole numbers less than and equal to $n$. E.g. $n=5$ $5! = 5 \times 4 \times 3 \times 2 \times 1.$ Note: We define the factorial $0!=1$.

#### Worked Examples

###### Example 1

Calculate $7!$.

###### Solution

From the definition we have $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1.$ Hence $7! =5040.$

###### Example 2

Simplify $\dfrac{9!}{6!}$.

###### Solution

Firstly we can expand the factorials using the definition $\dfrac{9!}{6!} = \dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}.$ This can then be simplified by canceling terms on the top and bottom to give $\dfrac{9!}{6!} = 9 \times 8 \times 7.$ Therefore, $\dfrac{9!}{6!} = 504.$

###### Example 3

Simplify $\dfrac{16!}{13!\cdot 8!}$

###### Solution

As above start by expanding the factorials $\dfrac{16!}{13!\cdot 8!} = \dfrac{16 \times 15 \times \dots \times 3 \times 2 \times 1}{(13 \times 12 \times 11 \times \dots \times 3 \times 2 \times 1 )( 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}.$ Then cancel out any terms that appear on the top and the bottom of the fraction $\dfrac{16!}{13!\cdot 8!} = \dfrac{16 \times 15 \times 14}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}.$ This can be canceled further if we factorize the numerator $\dfrac{16!}{13!\cdot 8!} = \dfrac{8 \times 2 \times 5 \times 3 \times 2 \times 7}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}.$ Then finally we get \begin{align} \dfrac{16!}{13! \cdot 8!} &= \dfrac{1}{4 \times 3},\\ &=\dfrac{1}{12}. \end{align}

###### Example 4

Simplify $\dfrac{(n-1)!}{(n+1)!}.$

###### Solution

\begin{align} \dfrac{(n-1)!}{(n+1)!} &= \dfrac{(n-1) \times (n-2) \times \dots \times 2 \times 1}{ (n+1) \times n \times (n-1) \times \dots \times 2 \times 1},\\ &= \dfrac{1}{(n+1) \times n},\\ &= \dfrac{1}{n^2 + n}. \end{align}