Define and evaluate factorials, use the recursive relation n! = n(n-1)!, and simplify factorial expressions and fractions by unrolling and cancelling

What this dot point is asking

Factorials are the arithmetic engine of the whole combinatorics topic, so Extension 1 starts you here, before permutations and combinations. NESA wants three things from this dot point. First, you should know what the symbol $n!$ means and be able to evaluate it, including the special convention $0! = 1$. Second, you should understand the recursive rule $n! = n(n-1)!$, because it is the rule that makes every later manipulation work. Third, and most often examined, you should be fluent at simplifying expressions and fractions built from factorials by "unrolling" and cancelling, rather than by computing two enormous numbers and dividing. The whole point of the notation is to let you cancel before you multiply.

The answer

Where factorials come from: arranging objects in a row

Almost every factorial you meet in this course answers the same kind of question: in how many ways can a set of distinct objects be arranged in a row? Suppose five people line up to form a queue at a bus stop. Fill the queue one position at a time, and count the choices at each stage.

• The person at the front can be any of the $5$ people: $5$ choices.
• One person is now placed, so the second position can be filled by any of the remaining $4$: $4$ choices.
• Then $3$ remain for the third position, then $2$ for the fourth, and finally just $1$ person is left for the last position.

Because the choices are made one after another and each is free of the others, you multiply the counts together (this is the multiplication principle, formalised on the next page). The number of possible queues is

That descending product, $5 \times 4 \times 3 \times 2 \times 1$, is written $5!$ and read "five factorial". The same reasoning with $n$ people in a row gives $n \times (n-1) \times \dots \times 2 \times 1$ arrangements, which is exactly $n!$. So whenever a question asks for the number of ways to arrange $n$ distinct objects in a line, the answer is $n!$, and that is why the notation earns its own section at the start of the topic.

The definition of n factorial

For each whole number $n \ge 1$, the number $n!$ is the product of every whole number from $n$ down to $1$:

The values grow extremely fast. It is worth knowing the first several by heart, because they recur constantly:

Beyond these, $8! = 40\,320$, $9! = 362\,880$ and $10! = 3\,628\,800$. The domain of the factorial is the whole numbers $0, 1, 2, \dots$ only: there is no factorial of a negative number or of a fraction, which is why a calculator returns an error for, say, $(-3)!$ or $(2.5)!$.

Why 0! = 1

The convention $0! = 1$ looks arbitrary the first time you see it, and it cannot come from the descending-product rule, because there is nothing to multiply. There are two ways to see that it is the only sensible choice, and you should be able to give at least one.

The counting argument is the cleaner one. $n!$ counts the arrangements of $n$ objects in a row, so $0!$ should count the arrangements of $0$ objects, that is, the arrangements of the empty set. There is exactly one such arrangement: the empty row, where nothing is placed. So $0! = 1$, not $0$.

The pattern argument backs this up. Going down the table, each factorial is the next one divided by the top number: $4! = 5! \div 5$, $3! = 4! \div 4$, $2! = 3! \div 3$, $1! = 2! \div 2$. Continuing the pattern, $0! = 1! \div 1 = 1$. Defining $0!$ any other way would break every formula that uses factorials, including the permutation and combination formulas later in the topic, so the convention is forced on us by consistency.

The recursive rule: n! = n(n-1)!

The single most useful fact about factorials is that each one is the next number down times the factorial below it:

This is true because $n! = n \times \big[(n-1) \times (n-2) \times \dots \times 1\big]$, and the bracket is exactly $(n-1)!$. For example $8! = 8 \times 7!$ and $10! = 10 \times 9!$. Read forwards, the rule builds factorials up one step at a time, with $0! = 1$ as the starting value. Read backwards, it lets you peel a factor off the front of a factorial, and that is the key to all the simplifying that follows.

Unrolling to simplify factorial fractions

Almost every exam manipulation of factorials is a fraction in which most of the product cancels. The method is always the same: take the larger factorial and unroll it, one factor at a time, until you reach the smaller factorial, which then cancels top and bottom. You never need to evaluate either factorial in full.

Take $\dfrac{10!}{7!}$. The bottom is $7!$, so unroll the top down to $7!$:

Notice you wrote out only the three factors above $7$ and stopped. The number of leftover factors is the gap between the two starting numbers: from $10$ down to (but not including) $7$ is exactly $3$ factors, $10, 9, 8$.

The same move works with pronumerals, which is where Extension 1 most often takes it. To simplify $\dfrac{(n+2)!}{(n-1)!}$, unroll the top from $n+2$ down to $(n-1)!$:

Again the count of leftover factors is the gap between the top index and the bottom index: $(n+2) - (n-1) = 3$, so exactly $3$ factors survive, $(n+2)$, $(n+1)$ and $n$. The general pattern, which you will meet again as the permutation formula, is

Combining factorial fractions over a common factorial

When two factorial fractions are added or subtracted, do not find some huge common denominator by multiplying the factorials together. Instead use the recursive rule to write the smaller factorial in terms of the larger one, so the larger factorial is already the common denominator.

For a numeric example, take $\dfrac{1}{8!} + \dfrac{1}{10!}$. Since $10! = 10 \times 9 \times 8!$, the common denominator is $10!$, and $\dfrac{1}{8!} = \dfrac{10 \times 9}{10!} = \dfrac{90}{10!}$. Therefore

The algebraic version is the more common exam target. To combine $\dfrac{1}{n!} + \dfrac{1}{(n+1)!}$, note $(n+1)! = (n+1) \times n!$, so $(n+1)!$ is the common denominator and $\dfrac{1}{n!} = \dfrac{n+1}{(n+1)!}$. Hence

The numerator collapses to a short expression precisely because the recursive rule did the heavy lifting; reaching for the product $n! \times (n+1)!$ as a denominator would have made the algebra far messier for no benefit.

Trailing zeroes of a factorial

A neat application that NESA likes, because it tests understanding rather than button-pushing, is counting how many zeroes a factorial ends in. You cannot do it on a calculator beyond about $13!$, since the display runs out of digits, but a short argument settles it exactly.

Every trailing zero comes from a factor of $10$, and $10 = 2 \times 5$. In the product $n! = n \times (n-1) \times \dots \times 1$ there are always far more factors of $2$ than factors of $5$ (every second number is even, but only every fifth is a multiple of $5$). So each trailing zero is governed by a factor of $5$, and the number of trailing zeroes equals the number of factors of $5$ in the product.

To count the factors of $5$ in $n!$, count the multiples of $5$ up to $n$, then add the multiples of $25$ (each of which contributes a second $5$), then multiples of $125$, and so on:

For instance $25!$ has $\left\lfloor \frac{25}{5} \right\rfloor + \left\lfloor \frac{25}{25} \right\rfloor = 5 + 1 = 6$ trailing zeroes, and $100!$ has $20 + 4 + 0 = 24$. The square term matters: $25 = 5^2$ slips in an extra factor of $5$ that the first count alone would miss.

How exam questions ask about factorials

The wording is predictable, and each phrasing maps to one of the moves above.

• "Evaluate $\dfrac{12!}{9!}$ (or with a $3!$ in the denominator) without a calculator." Unroll the larger factorial down to the smaller one and cancel; never compute the full factorials. If a factorial like $3!$ is left over, evaluate just that small piece.
• "Simplify $\dfrac{(n+2)!}{(n-1)!}$" or "write as a product of factors". Unroll the top to the bottom factorial; the number of surviving factors is the gap between the two indices.
• "Write $\dfrac{1}{n!} \pm \dfrac{1}{(n+1)!}$ as a single fraction." Use $(n+1)! = (n+1)\,n!$ so the larger factorial is the common denominator; rewrite the smaller-denominator fraction and combine.
• "How many zeroes does $n!$ end in?" Count the factors of $5$ with the floor-sum, remembering the extra $5$ from multiples of $25$.
• "Solve $\dfrac{(n+2)!}{n!} = 30$" or "$\dfrac{n!}{(n-2)!} = 56$". Unroll to turn the left side into a short product, form a quadratic, solve, and reject any root that is not a whole number $\ge 0$.
• "Show that $k \times k! = (k+1)! - k!$, hence sum a series." Take out the common factor $k!$ on the right to prove the identity, then use it to telescope the sum.