Factorial notation

In a nutshell

Using Pascal's triangle to expand $(x+y)^n$ can be inefficient. It is instead far quicker to use combinations; these often involve many terms which look like $m \times (m-1) \times (m-2) \times \dots \times 2 \times 1$ ; factorial notation provides a more compact way of writing such expressions.

Definition

For any positive whole number $n$ :

$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$

You pronounce $n!$ as " $n$ factorial". By definition, $0! = 1$ .

Note: For any positive whole number $n$ , $n!$ equals the number of ways of ordering a collection of $n$ objects - you have $n$ choices for the first object, then $n-1$ choices for the second, and so on. This gives $n \times (n-1) \times \dots \times 1 = n!$ orderings.

Curiosity: The "most sensible" answer to the question of how many ways you can order a collection of $0$ things is widely considered to be $1$ , which partly justifies defining $0! = 1$ .

Example 1

Compute $3!$

Using the above definition:

$3! = 3 \times 2 \times 1 = 6$

Therefore, $\underline { 3! = 6}$ .

Choosing r objects from n

The number of ways of choosing a collection of $r$ objects from a larger collection of $n$ objects is written as $^nC_r$ (or sometimes $\binom{n}{r}$ ). This is pronounced " $n$ choose $r$ ", and is defined in the following way:

^nC_r

$n$	A non-negative whole number; the number of objects being chosen from.
$r$	A non-negative whole number which is at most $n$ ; the number of objects being chosen.
$C$	Stands for "choose"; $^nC_r$ is the number of ways of choosing $r$ objects from a collection of $n$ .

Example 2

What is $^3C_1$ ?

$^3C_1$ denotes the number of ways of choosing $1$ object from a collection of $3$ objects.

Labelling these objects from $1$ to $3$ , you have $3$ choices - either pick object $1$ , or object $2$ , or object $3$ .

Therefore, $\underline{ ^3C_1 = 3}$ .

Computing $^nC_r$

Imagine choosing a collection of $r$ objects from a larger collection of $n$ objects one by one. You have $n$ choices for the first object, $n-1$ for the second and so on - until you arrive at $n-r+1$ choices for the $r^{th}$ object. This gives $n \times (n-1) \times (n-2) \times \dots \times (n-r+1)$ choices. More compactly,

$n \times (n-1) \times \dots \times (n-r+2) \times (n-r+1) = \dfrac{n!}{(n-r)!}$

However, you could have chosen those same $r$ objects in any order. There are $r!$ ways to order any collection of $r$ objects, so the above formula has overcounted by a factor of $r!$ Therefore:

$\boxed{^nC_r = \dfrac{n!}{r!(n-r)!}}$

Note: Most calculators come with buttons both for factorials and for inputting $^nC_r$ directly, making values of $^nC_r$ quick to compute if you have one at hand.

Example 3

What is $^6C_2$ ?

From the formula dervied above, you have that:

$^6C_2 = \dfrac{6!}{2!(6-2)!} = \dfrac{6!}{2!\times 4!}$ 

Compute:

$2! = 2 \times 1 = 2$

$4! = 4\times 3 \times 2 \times 1 = 24$ 

$6! =6\times 5 \times 4 \times 3 \times 2 \times 1 = 720$ 

Substitute these back in:

$\dfrac{6!}{2! \times 4!} = \dfrac{720}{2 \times 24} = 15$ 

Therefore, $\underline{^6C_2 = 15}$ .

Connection to Pascal's triangle

The following two statements about values of $^nC_r$ turns out to hold true for any positive whole numbers $r$ and $n$ with $r < n$ :

$^{n-1}C_{r-1} + \ ^{n-1}C_r = \ ^nC_r$

$^{n-1}C_0 = \ ^{n-1}C_{n-1} = 1$

These are the very same rules that you use to compute the $n^{th}$ row of Pascal's triangle from the row above; using this, it is possible to show that the $r^{th}$ entry of the $n^{th}$ row of Pascal's triangle is in fact equal to $^{n-1}C_{r-1}$ .

Example 4

Without using a calculator, work out the $7^{th}$ entry of the $10^{th}$ row of Pascal's triangle.

The $7^{th}$ entry of the $10^{th}$ row of Pascal's triangle is equal to:

$^{10-1}C_{7-1} = \ ^9C_6$

Substitute $9$ and $6$ into the formula for $^nC_r$ :

$^9C_6 = \dfrac{9!}{6!(9-3)!} = \dfrac{9!}{3! \times 6!}$

Simplify the expression:

$\dfrac{9!}{3! \times 6!} = \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$ 

Therefore the $7^{th}$ entry of the $10^{th}$ row of Pascal's triangle is $\underline{84}$ .