Factorial notation and permutations
Master factorial notation and permutations for Year 11 Specialist Mathematics in Queensland (QCAA). A factorial multiplies the whole numbers from a value down to one, and a permutation counts the ordered ways to arrange objects in a row.
You will learn to evaluate factorials, understand why order matters, and use the permutation formula to arrange some or all of a set of distinct objects — the counting skills behind combinations and probability later in the course.
Theory
Factorial notation and permutations are the tools for counting ordered arrangements in Year 11 Specialist Mathematics (QCAA, Queensland). A factorial \(n!\) multiplies the whole numbers from \(n\) down to \(1\), and \({}^{n}P_{r}=\dfrac{n!}{(n-r)!}\) counts the ways to arrange \(r\) of \(n\) distinct objects in order. This page shows how to evaluate factorials and permutations, with full worked examples.
A permutation is an ordered arrangement of objects: the order matters, so \(AB\) and \(BA\) are counted as two different arrangements. Counting permutations is about counting how many ways objects can be placed in a sequence.
Factorial notation gives a shorthand for the product of the descending whole numbers. For a positive integer \(n\), the factorial \(n!\) is \(n\times(n-1)\times\cdots\times2\times1\). By definition \(0!=1\).
To arrange all \(n\) distinct objects in a row you fill \(n\) positions with one fewer object available each time, giving \(n!\) arrangements. To arrange only \(r\) of the \(n\) objects, you use the permutation \({}^{n}P_{r}\).
The permutation \({}^{n}P_{r}\) counts the ordered ways to choose and arrange \(r\) of \(n\) distinct objects. It is a falling product of \(r\) factors starting at \(n\): \(n\times(n-1)\times\cdots\), which can be written compactly with factorials.
For a positive integer \(n\), the factorial is the descending product
The number of permutations of \(r\) objects chosen from \(n\) distinct objects is
How to evaluate a permutation
- Identify \(n\) (the number of distinct objects) and \(r\) (how many are being arranged in order).
- Write \({}^{n}P_{r}=\dfrac{n!}{(n-r)!}\), or start the falling product \(n\times(n-1)\times\cdots\) with exactly \(r\) factors.
- Cancel the common factorial (or stop the product after \(r\) factors) so only \(r\) terms remain.
- Multiply the remaining factors, one step at a time, to reach the answer.
A factorial multiplies the whole numbers from \(5\) down to \(1\); multiply one pair at a time:
| \(5!\) | \(=\) | \(5 \times 4 \times 3 \times 2 \times 1\) |
| \(=\) | \(20 \times 3 \times 2 \times 1\) | |
| \(=\) | \(60 \times 2 \times 1\) | |
| \(=\) | \(120 \times 1\) | |
| \(=\) | \(120\) |
\(5!=120\).
Use \({}^{n}P_{r}=\dfrac{n!}{(n-r)!}\) with \(n=8\), \(r=2\), then cancel the common factorial:
| \({}^{8}P_{2}\) | \(=\) | \(\dfrac{8!}{(8-2)!}\) |
| \(=\) | \(\dfrac{8!}{6!}\) | |
| \(=\) | \(\dfrac{8 \times 7 \times 6!}{6!}\) | |
| \(=\) | \(8 \times 7\) | |
| \(=\) | \(56\) |
\({}^{8}P_{2}=56\).
Arranging \(3\) of \(7\): with no repeats the choices fall \(7,6,5\) (see the slot diagram). Write it as a permutation:
| \({}^{7}P_{3}\) | \(=\) | \(\dfrac{7!}{(7-3)!}\) |
| \(=\) | \(\dfrac{7!}{4!}\) | |
| \(=\) | \(7 \times 6 \times 5\) | |
| \(=\) | \(42 \times 5\) | |
| \(=\) | \(210\) |
There are \(210\) ways.
Write \({}^{n}P_{2}\) as a falling product, form a quadratic, then factorise and solve:
| \({}^{n}P_{2}\) | \(=\) | \(n(n-1)\) |
| \(n(n-1)\) | \(=\) | \(56\) |
| \(n^2 - n\) | \(=\) | \(56\) |
| \(n^2 - n - 56\) | \(=\) | \(0\) |
| \((n-8)(n+7)\) | \(=\) | \(0\) |
| \(n\) | \(=\) | \(8\) |
\(n=8\) (reject \(n=-7\), since \(n\) is a positive integer).
Common pitfalls
Frequently asked questions
What does the factorial n! mean?
The factorial \(n!\) is the product of every whole number from \(n\) down to \(1\), that is \(n\times(n-1)\times\cdots\times2\times1\). For example \(5!=120\).
What is a permutation?
A permutation is an ordered arrangement of objects. Order matters, so \(AB\) and \(BA\) count as two different permutations.
What is the formula for nPr?
The number of permutations of \(r\) objects chosen from \(n\) distinct objects is \({}^{n}P_{r}=\dfrac{n!}{(n-r)!}=n\times(n-1)\times\cdots\times(n-r+1)\), a falling product of \(r\) factors.
Why does 0! equal 1?
By definition \(0!=1\). It is the value that makes \({}^{n}P_{n}=\dfrac{n!}{0!}=n!\), the number of ways to arrange all \(n\) objects, come out correctly.
How do you solve an equation like nP2 = 56?
Write \({}^{n}P_{2}=n(n-1)\), set it equal to \(56\), rearrange to the quadratic \(n^2-n-56=0\), factorise to \((n-8)(n+7)=0\), and take the positive integer \(n=8\).
What is the difference between a permutation and a combination?
A permutation counts ordered arrangements (order matters, \({}^{n}P_{r}\)); a combination counts unordered selections (order does not matter). This page covers permutations only.