Factorials

What you will be able to do

• write a product of whole numbers counting down to $1$ as a factorial;
• use the rule $n! = n \times (n - 1)!$, and explain why $0! = 1$;
• simplify ratios of factorials;
• write a product counting down from $n$ as a ratio of factorials.

A product that keeps appearing

Mira puts $5$ different books in a row on a shelf. The first place has $5$ choices, the next $4$, then $3$, $2$ and $1$. So there are

arrangements. Products like this one, counting down to $1$, appear in counting again and again. They get a short name.

The factorial

For a natural number $n$, the product $1 \times 2 \times 3 \times \dots \times n$ is written $n!$, and read "n factorial".

Each factorial is the one before it, times the new number:

For example, $6! = 6 \times 5! = 6 \times 120 = 720$. So you never need to multiply a long product from the start.

Why 0! = 1

The rule $n! = n \times (n - 1)!$ with $n = 1$ reads $1! = 1 \times 0!$. Since $1! = 1$, the only value that keeps the rule true is $0! = 1$. So we define $0! = 1$.

There is a counting reason too. There is exactly one way to arrange no objects: do nothing. Formulas in the next courses will give $0!$ in their answers, and $0! = 1$ makes them work every time.

Working with factorials

Factorials grow fast, so do not multiply them out. Cancel first.

• $\dfrac{10!}{8!} = \dfrac{10 \times 9 \times 8!}{8!} = 10 \times 9 = 90$.
• $\dfrac{7!}{5! \times 2!} = \dfrac{7 \times 6}{2} = 21$.

The reverse is useful as well. A product counting down from $n$ is a ratio of factorials:

The bottom, $6!$, cancels the numbers from $6$ down to $1$ that the product leaves out.

Some rules that look natural are false. In general $(2n)! \neq 2 \times n!$, and $(a + b)! \neq a! + b!$. For example, $4! = 24$, but $2 \times 2! = 4$.

A worked example

Find $x$ if $\dfrac{1}{7!} + \dfrac{1}{8!} = \dfrac{x}{9!}$.

1. Clear the fractions. Multiply every term by $9!$: $\dfrac{9!}{7!} + \dfrac{9!}{8!} = x$.
2. Cancel each ratio. $\dfrac{9!}{7!} = 9 \times 8 = 72$ and $\dfrac{9!}{8!} = 9$.
3. Add. $x = 72 + 9 = 81$.

Practise with new numbers

Try it in Python

The program builds each factorial from the one before, with the rule $n! = n \times (n - 1)!$. Then it compares with math.factorial, and cancels a ratio.

A look ahead

Mira now chooses only $3$ of her $5$ books, and puts them in a row. There are $5 \times 4 \times 3 = 60$ arrangements, which is $\dfrac{5!}{2!}$. Arrangements like this are called permutations, and the next course gives them their own formula.

Common mistakes

• Taking $0!$ as $0$. $0! = 1$. Otherwise the rule $1! = 1 \times 0!$ would fail.
• Cancelling the numbers inside the factorial. $\dfrac{6!}{3!}$ is $6 \times 5 \times 4 = 120$, not $2!$.
• Splitting a factorial over a sum or a product. $(a + b)!$ is not $a! + b!$, and $(2n)!$ is not $2 \times n!$.
• Multiplying out before cancelling. Cancel the common factorial first; the numbers stay small.

Remember these

• $n! = 1 \times 2 \times \dots \times n$, and $0! = 1$.
• $n! = n \times (n - 1)!$.
• $\dfrac{n!}{(n - 2)!} = n(n - 1)$, and the like: cancel the smaller factorial.
• A product of $k$ numbers counting down from $n$ is $\dfrac{n!}{(n - k)!}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $5! - 3!$.

2. Simplify $\dfrac{12!}{10!}$ without a calculator.

3. Write $11 \times 10 \times 9 \times 8$ as a ratio of factorials.

4. Is $3! + 3! = 6!$? Is $3! \times 4 = 4!$?

5. Find $n$ if $\dfrac{n!}{(n - 1)!} = 15$.