UK Secondary (7-11)
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Evaluate Factorial Expressions

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×(n1)×(n2)×(n3)×...×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!(n1)! can be simplified to $\frac{n\times\left(n-1\right)!}{\left(n-1\right)!}=n$n×(n1)!(n1)!=n.

Similarly, $\frac{n!}{\left(n-3\right)!}=n\times\left(n-1\right)\times\left(n-2\right)$n!(n3)!=n×(n1)×(n2), 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×(n3)! for $n$n, we might proceed as follows:

$n!$n! $=$= $120\times\left(n-3\right)!$120×(n3)!
$n\times\left(n-1\right)\times\left(n-2\right)\times\left(n-3\right)!$n×(n1)×(n2)×(n3)! $=$= $120\times\left(n-3\right)!$120×(n3)!
$n\times\left(n-1\right)\times\left(n-2\right)$n×(n1)×(n2) $=$= $120$120
$n^3-3n^2+2n$n33n2+2n $=$= $120$120
$n^3-3n^2+2n-120$n33n2+2n120 $=$= $0$0
$\left(n-6\right)\left(n^2+3n-120\right)$(n6)(n2+3n120) $=$= $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.



Question 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. 


Question 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 standard form on hand held calculators, and over $70!$70! won't compute on hand held calculators. 

Question 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

More Worked Examples


Simplify $\frac{7!}{5!}$7!5!


Simplify $\frac{6!}{4!4!}$6!4!4! by identifying common factors of the numerator and denominator.

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