8.04 Factorial notation (Enrichment)

What are factorials?

In 1808, the French mathematician Christian Kramp wrote "I use the very simple notation $n!$n! to designate the product of numbers decreasing from $n$n to unity, i.e. $n!=n\times\left(n-1\right)\times\left(n-2\right)\times\left(n-3\right)\times...\times3\times2\times1$n!=n×(n−1)×(n−2)×(n−3)×...×3×2×1".

Today we refer to the product as "$n$n factorial". So $5!=5\times4\times3\times2\times1=120$5!=5×4×3×2×1=120 and $6!=6\times5\times4\times3\times2\times1=720$6!=6×5×4×3×2×1=720. 

Factorials grow quickly. For example, $18!=6402373705728000$18!=6402373705728000. The number $100!$100! has $24$24 zeros at the right hand end of the number.  In fact, the number $59!$59! is larger than the current estimate of atoms in the universe!

Simplifying expressions involving factorials

Since $8!$8! is $8\times7!$8×7!, then an expression like $8!+7!$8!+7! can be simplified as $8\times7!+7!=7!\left(8+1\right)=9\times7!$8×7!+7!=7!(8+1)=9×7!. Similarly, $\frac{8!}{7!}=\frac{8\times7!}{7!}=8$8!7!​=8×7!7!​=8. 

More generally, an expression like $\frac{n!}{\left(n-1\right)!}$n!(n−1)!​ can be simplified to $\frac{n\times\left(n-1\right)!}{\left(n-1\right)!}=n$n×(n−1)!(n−1)!​=n.

Similarly, $\frac{n!}{\left(n-3\right)!}=n\times\left(n-1\right)\times\left(n-2\right)$n!(n−3)!​=n×(n−1)×(n−2), and this type of simplification can be useful in solving certain equations. For example, to solve $n!=120\times\left(n-3\right)!$n!=120×(n−3)! for $n$n, we might proceed as follows:

$n!$n!	$=$=	$120\times\left(n-3\right)!$120×(n−3)!
$n\times\left(n-1\right)\times\left(n-2\right)\times\left(n-3\right)!$n×(n−1)×(n−2)×(n−3)!	$=$=	$120\times\left(n-3\right)!$120×(n−3)!
$n\times\left(n-1\right)\times\left(n-2\right)$n×(n−1)×(n−2)	$=$=	$120$120
$n^3-3n^2+2n$n3−3n2+2n	$=$=	$120$120
$n^3-3n^2+2n-120$n3−3n2+2n−120	$=$=	$0$0
$\left(n-6\right)\left(n^2+3n-120\right)$(n−6)(n2+3n−120)	$=$=	$0$0

The factorisation of the cubic equation was completed using polynomial division techniques. The quadratic factor cannot be broken into linear factors over the reals. Hence $n=6$n=6. However, since $n$n is an integer, you might notice, at the third line, that $6\times5\times4=120$6×5×4=120, and so the solution is determined as $n=6$n=6 even before the factorisation step.

By definition, we say that $0!=1$0!=1.

 

Worked examples

example 1

Evaluate $5!$5!

This is a simple one, either on your calculator use the factorial button.  On my calculator it is listed as $x!$x! you may need to check your manual for yours, OR, we write down and evaluate long hand. 

$5!=5\times4\times3\times2\times1=120$5!=5×4×3×2×1=120

example 2

Evaluate $\frac{6!}{4!}$6!4!​

Long hand this looks like this 

$\frac{6!}{4!}$6!4!​

$=$=

$\frac{6\times5\times4\times3\times2\times1}{4\times3\times2\times1}$6×5×4×3×2×14×3×2×1​

But let's just write that right hand side a slightly different way. See how $6!$6! is also equal to $6\times5\times4!$6×5×4! So, 

$\frac{6!}{4!}$6!4!​	$=$=	$\frac{6\times5\times4!}{4!}$6×5×4!4!​
	$=$=	$6\times5$6×5
	$=$=	$30$30

Being able to simplify factorials in this way is very important, because anything higher than $12!$12! often results in scientific notation on hand held calculators, and over $70!$70! won't compute on hand held calculators. 

example 3

Evaluate $\frac{7!\times4!}{5!\times2!}$7!×4!5!×2!​

Let's use the trick we just learnt in question 2.  And simplify the numerator where we can first. 

$\frac{7!\times4!}{5!\times2!}$7!×4!5!×2!​	$=$=	$\frac{(7\times6\times5!)\times(4\times3\times2!)}{5!\times2!}$(7×6×5!)×(4×3×2!)5!×2!​
	$=$=	$7\times6\times4\times3$7×6×4×3
	$=$=	$504$504

Practice questions

QUESTION 1

QUESTION 2