Permutation, Combination & Probability

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Permutations: Arrangements

⏱ 4 min read🧩 6 question types🎯 12 practice Q

The idea in one minute

A permutation is an arrangement in which order matters. $^{n}P_r$ arranges $r$ of $n$ distinct things; a word of $n$ letters has $n!$ arrangements, divided by the repeats.

What a permutation is

A permutation is an arrangement in which order matters. ABC and ACB use the same letters but count as two arrangements.

Seats in a row, ranks in a race, digits in a number — all are permutations.

The nPr formula

$^{n}P_r$ counts the arrangements of $r$ items taken from $n$ distinct items.

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

Here $n!$ ( $n$ factorial) means $n \times (n-1) \times \cdots \times 1$ , and $0! = 1$ .

Fill the slots: the first has $n$ choices, the second $n-1$ , the $r$ -th has $n - r + 1$ .

$^{5}P_2 = 5 \times 4 = 20$ . A chair and a secretary from 5 people: 20 ways.

Words with all different letters

All letters distinct: the arrangements are $n!$ .

LEAD has 4 letters, so $4! = 24$ arrangements. PROBLEM has 7 letters, so $7! = 5040$ .

Words with repeated letters

Repeats make some arrangements look identical. Divide them out.

BETTER has 6 letters with E twice and T twice: $\dfrac{6!}{2!\,2!} = 180$ .

Rule: Divide $n!$ by the factorial of each letter's repeat count.

SUCCESS: 7 letters, S thrice, C twice: $\dfrac{7!}{3!\,2!} = 420$ .

Some people together

Tie the people who must sit together into one block. Arrange the block with the rest, then arrange inside the block.

5 friends in a row with A and B together: the block plus 3 others make 4 units, so $4! \times 2! = 48$ .

Some people never together

Two wordings need two methods.

"All girls sit together" uses the block method above.

"No two girls sit together" uses the gap method. Arrange the others first, then place the girls in the gaps.

4 boys and 3 girls, no two girls together: arrange the 4 boys, $4! = 24$ . Five gaps lie between and around them. Choose and fill 3 gaps: $^{5}P_3 = 60$ . Total $24 \times 60 = 1440$ .

Watch: "Total $-$ together" answers "at least two are together", not "no two are together".

Fixed places

Pin the people whose seats are fixed. Arrange only the free people.

6 people in a row, Ravi always first: $5! = 120$ .

Vowel-place questions work the same way. Place the vowels in their allowed spots first, then the consonants. STRANGE with vowels in the odd places: $^{4}P_2 \times 5! = 12 \times 120 = 1440$ .

Round tables and necklaces

Around a round table, rotations are the same seating. Fix one person and arrange the rest.

\text{round table} = (n-1)!

5 people: $4! = 24$ ways.

In a necklace or garland, the flipped arrangement looks the same too. Divide by 2 more: $\dfrac{(n-1)!}{2}$ .

Note: Necklaces, garlands and key chains divide by 2. Seating people does not — a mirror image around a table is a different seating.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Arrange r of n (nPr)

How to spot it:

'In how many ways can r of n things be arranged', or ordered posts like president and secretary are filled.

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

Method

Check that order matters.
Compute $^{n}P_r = n!/(n-r)!$ .
Expand only as far as needed: $^{7}P_3 = 7 \times 6 \times 5$ .

Why it works:

Filling $r$ ordered slots from $n$ items multiplies $n$ , $n-1$ , … down to $n-r+1$ .

Try this

A chairperson and a secretary are chosen from 8 members. In how many ways can this be done?

Show solution

Two ordered posts: $^{8}P_2$ .
$8 \times 7 = 56$ .

Answer

Type 2very common2 practice Q

Arrange all letters of a word (all distinct)

How to spot it:

'Arrange the letters of the word…' and every letter is different.

n!

Method

Count the letters.
All distinct means the answer is the factorial of the count.
Use remembered factorials up to $7!$ .

Why it works:

Every ordering of all $n$ letters is a full permutation, and there are $n!$ of them.

Try this

In how many ways can the letters of LEAD be arranged?

Show solution

4 distinct letters.
$4! = 24$ .

Answer

Type 3very common2 practice Q

Arrange letters of a word with repeats

How to spot it:

'Arrange the letters of…' and a letter appears twice or more (LETTERS, SUCCESS, BANANA).

\frac{n!}{p!\,q!\,r!}

Method

Count all the letters ( $n$ ).
Note each letter's repeat count.
Divide $n!$ by the factorial of each repeat count.

Why it works:

Swapping identical letters changes nothing, so each distinct word is counted $p!\,q!\ldots$ times.

Try this

In how many ways can the letters of BETTER be arranged?

Show solution

6 letters; E twice, T twice.
$\dfrac{6!}{2!\,2!} = 180$ .

Answer

180

Type 4very common2 practice Q

Restrictions: together or never together

How to spot it:

'Two people always sit together' (block method) or 'no two girls sit together' (gap method).

Method

Together: tie them into one block; arrange units, then arrange inside the block.
Never together: arrange the others first.
Place the restricted people in the gaps between the others.
Multiply the stage counts.

Why it works:

A block acts as one unit, and gaps are the only seats that keep people apart.

Try this

In how many ways can 5 friends sit in a row if two particular friends always sit together?

Show solution

Block + 3 others = 4 units: $4!$ .
Inside the block: $2!$ .
$24 \times 2 = 48$ .

Answer

Type 5common2 practice Q

Fixed places for particular people

How to spot it:

'A always sits in the first seat', 'the vowels occupy the odd places'.

Method

Fix the pinned people or letters first.
Count the ways to place them in their allowed spots.
Arrange the remaining people in the remaining spots.
Multiply the two counts.

Why it works:

Fixed seats remove people from the pool, and the rest arrange freely.

Try this

In how many ways can 6 people sit in a row if Ravi always sits in the first seat?

Show solution

Ravi is fixed: 1 way.
The rest: $5! = 120$ .

Answer

120

Type 6common2 practice Q

Circular seating and necklaces

How to spot it:

'Around a round table' or 'in a circle' — or beads, flowers, necklace, garland.

(n-1)!; \quad \text{necklace} = \frac{(n-1)!}{2}

Method

For a round table, fix one person to kill the rotations.
Arrange the other $n-1$ people: the answer is $(n-1)!$ .
For a necklace or garland, also divide by 2 for the flip.

Why it works:

Rotating everyone gives the same circular seating, so one fixed anchor removes the overcount.

Try this

In how many ways can 5 people be seated around a round table?

Show solution

Fix one person.
$4! = 24$ .

Answer

Formula sheet

Arrange r of n

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

Order matters; $0! = 1$.

All letters of a word

n!

All $n$ letters distinct.

Word with repeated letters

\frac{n!}{p!\,q!\,r!}

$p$, $q$, $r$ are the repeat counts of the repeated letters.

Round table

(n-1)!

Fix one person; rotations are the same seating.

Necklace or garland

\frac{(n-1)!}{2}

Flips look identical, so divide by 2.

Shortcuts that save time

⚡ Block method for 'together'

Tie the must-together people into one bundle. Arrange the bundle with the others, then arrange inside the bundle.

Example

In how many ways can A, B, C, D, E sit in a row with A and B together?

Show solution

4 units: $4! = 24$ .
Inside the block: $2$ .
$24 \times 2 = 48$ .

Answer

⚡ Divide out the repeats

Repeated letters divide the factorial. Learn the repeat patterns of common exam words as you practise.

Example

How many arrangements does ALL have?

Show solution

3 letters, L twice.
$\dfrac{3!}{2!} = 3$ .

Answer

⚡ Fix one person for circles

In a circle, fix one person and arrange the rest. For necklaces and garlands, divide by 2 more.

Example

In how many ways can 6 people sit around a round table?

Show solution

Fix one person.
$5! = 120$ .

Answer

120

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $n!$ when letters repeat — BETTER has $\dfrac{6!}{2!\,2!} = 180$ arrangements, not $720$ .

Mistake 02

Forgetting the $2!$ inside the block — 'A and B together' is $4! \times 2!$ , not just $4!$ .

Mistake 03

Answering 'total $-$ together' for 'no two girls sit together' — that removes only the all-together case; use the gap method.

Mistake 04

Giving a round table $n!$ seatings — rotations are the same seating, so it is $(n-1)!$ .

Mistake 05

Dividing a necklace answer by 2 and forgetting, or dividing a round-table answer by 2 when it should not be.

Quick revision

Read this the night before the exam.

Order matters, so it is a permutation.
$^{n}P_r = \dfrac{n!}{(n-r)!}$ ; all letters of a word: $n!$ .
Repeated letters: divide by each repeat's factorial.
Together: (units)! $\times$ (inside block)!.
Never together: arrange the rest, then fill the gaps.
Round table $(n-1)!$ ; necklace $\dfrac{(n-1)!}{2}$ .

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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