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.

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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:

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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.

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Example 3

What is $$^6C_2$$?

From the formula dervied above, you have that:

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 $$\underline{84}$$.

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