Permutations and Combinations

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Evaluate Factorial Expressions

Lesson

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!

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.

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

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.

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 |

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

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

Use permutations and combinations