Permutations: Counting Arrangements

Count arrangements where order matters using n! and nPr, handle restrictions, items together or apart, and solve equations with nPr.

In the exam for IGCSE Additional Maths 0606 · 13 real past-paper questions below

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Filling places one by one

In how many different orders can you put 4 different books on a shelf? You could list them all, but there is a much faster way.

Fill the places one at a time. The first place can take any of the 4 books. Once it is filled, 3 books are left for the second place, then 2, then 1.

Worked example

1
Count the orders for 4 different books on a shelf.
2
$4 \times 3 \times 2 \times 1$ — Choices for place 1, then place 2, then 3, then 4.
3
$= 24$ — There are 24 different orders.

Rule

When you make one choice and then another, multiply the numbers of choices.

An ordering like this is called a permutation. In a permutation, order matters: Book A then Book B is different from Book B then Book A.

Multiplying all the whole numbers down to 1 happens so often that it has its own symbol, the factorial.

Formula

n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1

$n$ — the number of different items you arrange.
$n!$ — say “n factorial”. It counts the ways to arrange all n items in a row.

Rule

By definition $0! = 1$ . It follows the pattern $n! = n \times (n - 1)!$ : put $n = 1$ and you get $1! = 1 \times 0!$ , so $0!$ must be 1.

Key idea

A pack of 52 playing cards can be shuffled into 52! orders. That is about 8 followed by 67 zeros, so a well-shuffled pack is almost surely in an order no pack has ever been in before.

Key idea

Count places, not lists. Multiply the number of choices for each place.

Arranging r out of n

Often you only arrange some of the items. For example, a 3-letter code using 3 different letters from 8. You fill only 3 places, so you stop multiplying after 3 numbers.

Worked example

1
How many 3-letter codes can be made from 8 different letters, using each at most once?
2
$8 \times 7 \times 6$ — 8 choices for the first place, 7 for the second, 6 for the third.
3
$= 336$ — Stop after 3 places.

This has its own notation. The number of permutations of r items taken from n is written $^{n} P_{r}$ .

Formula

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

$n$ — how many different items you can choose from.
$r$ — how many places you fill, in order.
$(n - r)!$ — cancels the factors you do not use, so only r numbers are multiplied.

Rule

This formula is not on the 0606 or 4037 formula list. Learn it. Your calculator has an nPr key for Paper 2.

Worked example

1
Find $^{8} P_{5}$ , the number of ways to arrange 5 of 8 different letters in a row.
2
$^{8} P_{5} = \frac{8 !}{3 !}$ — Here n = 8 and r = 5, so n − r = 3.
3
$= 8 \times 7 \times 6 \times 5 \times 4$ — The 3 × 2 × 1 cancels. Five numbers are left, one for each place.
4
$= 6720$ — Multiply.

Common mistake

Writing $\frac{n !}{r !}$ . The bottom is $(n - r)!$ , the factorial of the items you do not use.

Key idea

$^{n} P_{r}$ is just r numbers multiplied, counting down from n.

Restrictions: fill the fussy place first

Many questions add a condition: the number must be even, must start with a certain digit, or must be greater than 60 000. Deal with the restricted place first, then fill the rest.

Worked example

1
How many 5-digit numbers greater than 60 000 can be made from the digits 1 to 8, with no digit repeated?
2
$First digit: 6, 7 or 8$ — 3 choices. Any of these makes the number bigger than 60 000.
3
$7 \times 6 \times 5 \times 4 = 840$ — 7 digits are left for the other 4 places, in order.
4
$3 \times 840 = 2520$ — Multiply the choices.

If one choice changes the number of choices somewhere else, split the question into cases. Count each case, then add.

Worked example

1
Using the digits 0, 1, 2, 3, 4, 5, 6 with no repeats, how many 4-digit even numbers can be made? A number cannot start with 0.
2
$Case 1: ends in 0$ — Then 0 is used up, so the first place has no problem.
3
$6 \times 5 \times 4 = 120$ — 6 digits are left for the first three places.
4
$Case 2: ends in 2, 4 or 6$ — 3 choices for the last digit.
5
$3 \times 5 \times 5 \times 4 = 300$ — First place: not 0 and not the last digit, so 5 choices. Then 5 and 4 for the middle.
6
$120 + 300 = 420$ — Add the cases.

The mark scheme gives a method mark for each correct case. Write each case on its own line with its product.

Common mistake

Using $3 \times 6 \times 5 \times 4$ for every even ending. When the last digit is 0, the first place has more choices. The 0 changes things, so split into cases.

Common mistake

Counting a number twice in two cases. Make the cases separate: ends in 0, or ends in a non-zero even digit.

Key idea

Fill the fussy place first. If the fussy place changes another place, split into cases and add.

Together and apart

Some people or objects must stand together. Tie them into one block. Arrange the blocks, then arrange the items inside the block.

Worked example

1
4 boys and 3 girls stand in a line. How many arrangements have the 3 girls together?
2
$4 boys + 1 block = 5 items$ — The 3 girls act as one block.
3
$5! = 120$ — Arrange the 5 items.
4
$3! = 6$ — The girls can change order inside their block.
5
$120 \times 6 = 720$ — Multiply.

If no two girls may stand next to each other, use the gaps. Line up the boys first. The girls go into the gaps between and at the ends of the boys.

Worked example

1
Same group. How many arrangements have no two girls next to each other?
2
$4! = 24$ — Arrange the boys.
3
$□ B □ B □ B □ B □$ — 4 boys make 5 gaps, shown as boxes: one at each end and three between.
4
$^{5} P_{3} = 5 \times 4 \times 3 = 60$ — Put the 3 girls into 3 of the 5 gaps, in order.
5
$24 \times 60 = 1440$ — Multiply.

“Not all three together” is a different question. There, use everything minus the together arrangements: $7! - 720 = 4320$ .

Common mistake

Treating “no two together” and “not all together” as the same. No two together means every girl has a boy beside her. Not all together only rules out the full block of three.

Common mistake

Forgetting to arrange the items inside the block. Multiply by the factorial of the block size.

Key idea

Together: make a block, then arrange inside it. Apart: arrange the others, then use the gaps.

Equations with nPr

Some questions use permutations in algebra. On 0606 Paper 1 there is no calculator, so you must cancel the factorials by hand.

Rule

Write $^{n} P_{r}$ as a product of r factors counting down from n. For example, $^{n} P_{3} = n (n - 1) (n - 2)$ .

Worked example

1
Find n, given that $^{n} P_{4} = 20 \times^{n} P_{2}$ .
2
$n (n - 1) (n - 2) (n - 3) = 20 n (n - 1)$ — Write both sides as products.
3
$(n - 2) (n - 3) = 20$ — Divide both sides by n(n − 1). It is not 0, because n is at least 4.
4
$n^{2} - 5 n - 14 = 0$ — Expand and collect.
5
$(n - 7) (n + 2) = 0$ — Factorise.
6
$n = 7$ — Reject n = −2. You cannot have a negative number of items.

The marks go for writing the factorials correctly, then the simplified equation, then the answer. So show the product line before you simplify.

Common mistake

Keeping a negative or fraction answer. n counts items, so it must be a whole number, at least as big as r.

Key idea

Turn each $^{n} P_{r}$ into a short product, cancel the common factors, and you are left with an ordinary equation.

Which method? Read the question

Read the words first. They tell you which tool to pick.

You see	You do
“Arrange all n items”	$n!$
“Arrange r of the n items” or a code or number	$^{n} P_{r}$
“Must start with” or “must end with”	Fill that place first
“Even”, “divisible by 5” or “greater than” with a 0 in the digits	Split into cases, then add
“Must be together”	Make a block: arrange blocks × arrange inside
“No two next to each other”	Arrange the others, then use the gaps
“Not all together”	Total minus together
An equation with $^{n} P_{r}$	Write as products, cancel, solve, reject bad values

In every question here, order matters: a different order is a different answer. When order does not matter, you are choosing, not arranging. That is the combinations lesson.

Key idea

Picture the places in a row. Fill the fussy places first, then multiply.

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