MathsISC Class 11 · Permutations and Combinations

Permutations: ⁿPᵣ, Restricted Arrangements, Repeated Objects and Circular Cases

Updated 17 October 2026 · Fact-checked

The number of arrangements of r objects chosen from n distinct objects is ⁿPᵣ = n!/(n − r)!. If p objects are alike of one kind and q of another, the arrangements of all n are n!/(p! q!). For n distinct objects round a circle the number of arrangements is (n − 1)!. For restrictions "together" and "never together", glue the objects or use gaps.

Place the unrestricted people first, then choose separate gaps for the restricted ones: 3 boys make 4 gaps, so 3! × ⁴P₂ = 72 (same as 120 − 48).

Illustration drawn for QuestionBuddy.

Understand Permutations: Arrangements and Restricted Cases

Fundamental principle. If one task can be done in m ways and a second in n ways, the two together can be done in $m \times n$ ways (multiplication principle); if either, but not both, then $m + n$ ways (addition principle).

Permutations. $n! = n (n - 1) ...1$ with $0! = 1$ . $^{n} P_{r} = \frac{n !}{( n - r )!}$ is the number of ordered selections of r from n distinct objects. $^{n} P_{n} = n!$ . With repetition allowed, r positions can be filled in $n^{r}$ ways.

Number formation. For digits, fill the restricted position first (the first place cannot be 0; an even number needs an even last digit).

Alike objects. The arrangements of n objects of which $p_{1}$ are alike of one kind, $p_{2}$ alike of another and so on: $\frac{n !}{p _{1} ! p _{2} ! ...}$ . For MISSISSIPPI: $\frac{11 !}{4 ! 4 ! 2 !}$ .

Together and never together. To keep some objects together, treat them as a single object, arrange it with the others, then multiply by the arrangements within the group. For "never together" compute total − together, or place the other objects first and put the restricted objects in the gaps between them.

Circular permutations. n distinct objects round a circle can be arranged in $(n - 1)!$ ways (rotations are the same). If clockwise and anticlockwise arrangements are alike (necklace, garland), the number is $\frac{( n - 1 )!}{2}$ .

Permutations: Arrangements and Restricted Cases: formulas to remember

Permutations: $^nP_r = \frac{n!}{(n - r)!}$ | $^nP_n = n!$; Order matters; objects distinct.
Alike objects: $\frac{n!}{p! q! r!}$; For p, q, r alike of three kinds.
Circular: $$(n - 1)!$ | $\frac{(n - 1)!}{2}$ for necklaces$; Rotation does not give a new arrangement.
With repetition: $$n^r$ words of length r from n symbols$; Each place has n choices.

Permutations: Arrangements and Restricted Cases in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Seating at a round table

Suppose 7 friends sit round a circular table.

Find the number of arrangements, the number in which two particular friends sit together, and the number in which they do not sit together.

Try it first, then show the solutionSolution

Circular arrangements $= (7 - 1)! = 720$ .
Two together: treat them as one, giving 6 objects in $(6 - 1)! = 120$ arrangements, times 2 inside the pair: $240$ .
Not together $= 720 - 240 = 480$ .

Answer: 720; 240; 480.

Memory hook: Fix one object to account for rotations.

Tricks to remember Permutations: Arrangements and Restricted Cases

Short memory aids. Say each one aloud once, then use it on a question the same day.

Glue for together
Treat the group as one object, then multiply by the internal arrangements.
Use it when: Together arrangements.
Gaps for never together
Seat the unrestricted ones first and use the gaps between them.
Use it when: No two adjacent.

Permutations: Arrangements and Restricted Cases: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Value of ⁿPᵣ · 1 mark · MCQBoard level

The value of $^{8} P_{3}$ is:
(a) 56
(b) 336
(c) 512
(d) 40320

Try it first, then show the solutionSolution

$8 \times 7 \times 6 = 336$ .

Answer: (b) 336

Example 2: Number formation · 3 marks · Short answerBoard level

How many 4-digit numbers can be formed from the digits 1 to 9 without repetition? How many of them are even?

Try it first, then show the solutionSolution

Total $=^{9} P_{4} = 9 \times 8 \times 7 \times 6 = 3024$ . (1 mark)
Even: the last place takes one of 2, 4, 6, 8: 4 ways. (1 mark)
The other three places: $^{8} P_{3} = 336$ ; even numbers $= 4 \times 336 = 1344$ . (1 mark)

Answer: 3024; 1344.

Example 3: Repeated letters · 4 marks · Short answerHarder board level

Find the number of arrangements of the letters of MISSISSIPPI, and the number in which all four S are together.

Try it first, then show the solutionSolution

Letters: M1, I4, S4, P2; total $= \frac{11 !}{4 ! 4 ! 2 !} = \frac{39916800}{24 \times 24 \times 2} = 34650$ . (2 marks)
Treat SSSS as one object: 8 objects with I4, P2 alike: $\frac{8 !}{4 ! 2 !} = \frac{40320}{48} = 840$ . (2 marks)

Answer: 34650; 840.

Example 4: No two women together · 4 marks · Short answerHarder board level

Five men and four women are to sit in a row. Find the number of arrangements in which no two women are together.

Try it first, then show the solutionSolution

Seat the 5 men first: $5! = 120$ ways. (1 mark)
They create 6 gaps (including the ends) for the women. (1 mark)
Choose and arrange 4 women in 6 gaps: $^{6} P_{4} = 360$ . (1 mark)
Total $= 120 \times 360 = 43200$ . (1 mark)

Answer: 43200

Example 5: Circular arrangements · 5 marks · Long answerHOTS / competency

Seven beads of different colours are strung on a necklace. Find the number of necklaces. In how many of them are two particular beads adjacent?

Try it first, then show the solutionSolution

A necklace can be turned over, so the number is $\frac{( 7 - 1 )!}{2} = 360$ . (1.5 marks)
Two beads adjacent: treat them as one, so 6 objects: $\frac{( 6 - 1 )!}{2} = 60$ . (1.5 marks)
The pair can be arranged internally in 2 ways: $60 \times 2 = 120$ . (1 mark)
Check: adjacency in a circle of 7 is the same for each of the 6 positions of the pair; the answer 120 is one third of 360. (1 mark)

Answer: 360 necklaces; 120 with the two beads adjacent.

Where students lose marks in Permutations: Arrangements and Restricted Cases

The slips examiners see most often, and what to write instead.

Mistake: Forgetting the first-digit restriction
Allowing 0 as the first digit.
Do this: Fill the restricted place first.
Usually costs: 2 marks.
Mistake: Not dividing for alike objects
Using n! for repeated letters.
Do this: Divide by the factorials of the repeats.
Usually costs: 2 marks.
Mistake: Using n! for circular arrangements
Counting rotations separately.
Do this: Use (n − 1)!.
Usually costs: 1 to 2 marks.

Exam tips

State the principle you use (multiplication or addition).
Handle the restricted position first.
Check small cases mentally.
Write factorials before cancelling.

Practice questions on Permutations: Arrangements and Restricted Cases

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of arrangements of the letters of the word BOOK is:
(a) 24
(b) 12
(c) 6
(d) 4
Hint
Two O's alike.
Answer
(b) 12
Q2 · 2 marks · Short answerBoard level
How many 3-digit numbers can be formed from 0, 1, 2, 3, 4 without repetition?
Hint
The first place cannot be 0.
Answer
First digit: 4 ways; the next two: $^{4} P_{2} = 12$ ; total $4 \times 12 = 48$ .
Q3 · 3 marks · Short answerHarder board level
In how many ways can the letters of the word SUNDAY be arranged so that the vowels are not together?
Hint
Total minus together.
Answer
Total $6! = 720$ . Vowels U, A together: treat as one: $5! \times 2 = 240$ . Not together $= 480$ .
Q4 · 3 marks · Short answerHarder board level
How many 5-letter words, with or without meaning, can be formed from the letters of EQUATION if no letter is repeated and the word begins with a vowel?
Hint
EQUATION has 5 vowels.
Answer
First letter: 5 vowels; remaining 4 places from the other 7 letters: $^{7} P_{4} = 840$ ; total $5 \times 840 = 4200$ .
Q5 · 3 marks · Short answerHarder board level
Find r if $^{12} P_{r} = 1320$ .
Hint
12 × 11 × 10 = 1320.
Answer
$12 \times 11 \times 10 = 1320$ so $r = 3$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a bank gives a customer a 4-digit PIN using the digits 0 to 9.
(i) How many PINs are possible if digits may repeat? (1 mark)
(ii) How many have all digits different? (1 mark)
(iii) How many all-different PINs are even? (1 mark)
(iv) How many have the first digit non-zero and no repetition? (1 mark)
Hint
Use $n^{r}$ and $^{n} P_{r}$ .
Answer
(i) $1 0^{4} = 10000$ . (ii) $^{10} P_{4} = 5040$ . (iii) Last digit even (5 choices) and the other three from the remaining 9: $5 \times^{9} P_{3} = 5 \times 504 = 2520$ . (iv) $9 \times^{9} P_{3} = 9 \times 504 = 4536$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.ⁿPᵣ?Show answer
n!/(n − r)!.
2.Circular arrangements?Show answer
(n − 1)!.
3.0!?Show answer
1.
4.Arrangements with alike objects?Show answer
n!/(p! q! ...).

Permutations: Arrangements and Restricted Cases: frequently asked questions

When does order matter?

In permutations the order of selection matters; in combinations it does not.

Why divide by 2 for a necklace?

A necklace can be turned over, so an arrangement and its mirror image are the same.

What is the difference between "with repetition" and "without repetition"?

With repetition, the same symbol may be used again, giving n^r; without, we use ⁿPᵣ.