The factorial n factorial is the product of the whole numbers from n down to one. It counts the arrangements of n distinct objects in a row. Also zero factorial equals one.

Year 11 Methods (Unit 1 & 2) Counting Methods And Binomial Expansions

Arrangements

Q: What is the difference between a permutation and a combination?

A permutation counts ordered arrangements, so order matters. A combination counts unordered selections, where order does not matter.

Q: What does the formula for n P r mean?

n P r equals n factorial over open bracket n minus r close bracket factorial. It counts the ordered ways to place r objects in positions from n distinct objects, which equals n times n minus one and so on with r factors.

Q: How do I count arrangements with repeated letters?

Take n factorial for all the letters, then divide by the factorial of each repeated group, giving n factorial over p factorial q factorial and so on.

20 practice questions 1 video lesson Theory + worked examples

Explore arrangements, or permutations — a supporting counting idea alongside Queensland Year 11 Mathematical Methods (QCAA). An arrangement counts the ordered ways objects can be placed in a line, where order matters.

Because QCAA Methods counts only combinations, or unordered selections, arrangements are background rather than a listed topic. You will use factorials, arrange distinct objects in a row, and see how removing order turns an arrangement into a combination.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), an arrangement (or permutation) is an ordered listing of objects, where the order matters. This page covers factorials \(n!\), arranging \(n\) distinct objects in a row, the ordered-selection count \(^{n}P_{r}=\dfrac{n!}{(n-r)!}\), arrangements with identical items, and simple restrictions such as keeping two items together.

The factorial \(n!\) is the product of the whole numbers from \(n\) down to \(1\); for example \(5!=5\times4\times3\times2\times1=120\). It counts the number of ways to arrange \(n\) distinct objects in a row, and by convention \(0!=1\).

An arrangement of \(r\) objects chosen in order from \(n\) distinct objects is a permutation, written \(^{n}P_{r}\). Filling \(r\) positions gives \(n(n-1)\cdots\) with \(r\) factors, which equals \(\dfrac{n!}{(n-r)!}\).

When some objects are identical, divide by the factorial of each repeated group to remove the arrangements you cannot tell apart.

Order matters in an arrangement. Swapping two objects gives a different arrangement, which is why arrangements use factorials.

\(^{7}P_{3}=7\times6\times5=210\): three ordered positions filled from seven objects.

One ordered arrangement of \(4\) distinct tiles; there are \(4!=24\) in all.

Factorial and arranging \(n\) distinct objects in a row:

\[n!=n\times(n-1)\times\cdots\times2\times1\]

n! = n \times (n - 1) \times \dots \times 1

Permutations — ordered selections of \(r\) from \(n\):

\[^{n}P_{r}=\dfrac{n!}{(n-r)!}\]

{}_{n}P_{r} = \frac{n!}{(n - r)!}

Arrangements with identical items (groups of \(p,q,\dots\) alike):

\[N=\dfrac{n!}{p!\,q!\cdots}\]

N = \frac{n!}{p! q! \dots}

Keep-together trick: glue the items that must stay together into one block, arrange the blocks, then multiply by the internal arrangements of the block.

How to count arrangements

Decide the positions: how many objects are being placed in order, and how many positions there are.
Fill the boxes: use \(n!\) for all \(n\) in a row, or \(^{n}P_{r}=\dfrac{n!}{(n-r)!}\) for \(r\) of them in order.
Adjust: divide by \(p!\,q!\cdots\) for identical items, or glue a must-stay-together block and multiply by its internal arrangements.

Example 1 — A factorial

In how many ways can \(5\) different books be arranged in a row on a shelf?

Solution

Five distinct books fill five positions, so use \(5!\):

\(N\)	\(=\)	\(5!\)
	\(=\)	\(5\times 4\times 3\times 2\times 1\)

Multiply:

\(N\)

\(=\)

\(120\)

There are \(120\) arrangements.

5! = 120

Example 2 — A permutation \(^{n}P_{r}\)

From \(7\) different books, in how many ways can \(3\) be placed in order on a shelf?

Solution

Ordered selection of \(3\) from \(7\), so use \(^{7}P_{3}=\dfrac{7!}{(7-3)!}\):

\(^{7}P_{3}\)

\(=\)

\(\dfrac{7!}{4!}\)

Cancel the \(4!\) — only the top three factors survive:

	\(=\)	\(\dfrac{7\times 6\times 5\times 4!}{4!}\)
	\(=\)	\(7\times 6\times 5\)

Multiply:

\(^{7}P_{3}\)

\(=\)

\(210\)

There are \(210\) ordered arrangements.

{}_{7}P_{3} = 210

Example 3 — Identical letters

How many distinct arrangements are there of the letters of the word \(\text{BANANA}\)?

Solution

There are \(6\) letters with \(3\) A's and \(2\) N's repeated, so divide by \(3!\,2!\):

\(N\)

\(=\)

\(\dfrac{6!}{3!\,2!}\)

Evaluate the factorials:

	\(=\)	\(\dfrac{720}{6\times 2}\)
	\(=\)	\(\dfrac{720}{12}\)

Divide:

\(N\)

\(=\)

\(60\)

There are \(60\) distinct arrangements.

\frac{6!}{3! 2!} = 60

Example 4 — Keep two together

Five friends sit in a row of \(5\) seats. In how many ways can they sit if two particular friends must sit next to each other?

Solution

Glue the two friends into one block, leaving \(4\) items to arrange:

\(\text{blocks}\)	\(=\)	\(4!\)
	\(=\)	\(24\)

The two friends can swap within their block:

\(\text{inside}\)	\(=\)	\(2!\)
	\(=\)	\(2\)

Multiply the two counts together:

\(N\)	\(=\)	\(24\times 2\)
	\(=\)	\(48\)

There are \(48\) seatings.

4! \times 2! = 48

Common pitfalls

Using order when it does not matter. An arrangement counts order; if the question asks for a group or committee where order is irrelevant, that is a selection (combination), not a permutation.

Forgetting to divide for repeats. With identical letters, \(6!\) overcounts; divide by \(3!\,2!\) to remove the arrangements you cannot tell apart.

Dropping the internal swap. For a keep-together block, remember to multiply by the arrangements inside the block (\(2!\) for two items).

Frequently asked questions

What is a factorial?

The factorial \(n!\) is the product of the whole numbers from \(n\) down to \(1\); it counts the arrangements of \(n\) distinct objects in a row. Also \(0!=1\).

What is the difference between a permutation and a combination?

A permutation counts ordered arrangements, so order matters; a combination counts unordered selections, where order does not matter.

What does the formula for nPr mean?

\(^{n}P_{r}=\dfrac{n!}{(n-r)!}\) counts the ordered ways to place \(r\) objects in positions from \(n\) distinct objects; it equals \(n(n-1)\cdots\) with \(r\) factors.

How do I count arrangements with repeated letters?

Take \(n!\) for all the letters, then divide by the factorial of each repeated group, giving \(\dfrac{n!}{p!\,q!\cdots}\).

How do I keep two people together?

Treat the pair as a single block, arrange the blocks, then multiply by \(2!\) for the two ways they can sit within the block.

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