Permutations and Combinations Class 11: Notes, Formulae and Solved Questions

What Permutations and Combinations is about

Weightage: This chapter is part of Unit II, Algebra, which the syllabus gives 25 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit II Algebra

Key concepts

Fundamental principle of counting

NCERT section 6.2

Multiplication principle: if an event can happen in m different ways, and after it a second event can happen in n different ways, then the two events in that order can happen in $m\times n$ ways. It extends to any number of events: $m\times n\times p\times \dots$

How to use it: draw one box for each position (or stage), write in each box the number of choices for it, and multiply.

• With repetition allowed, every box usually gets the full number of choices.
• Without repetition, each box has one fewer choice than the one before.
• Fill any restricted box first (first digit cannot be 0; last digit must be even).

Worked example

How many 3-digit numbers can be formed from the digits 1, 2, …, 6 if (i) repetition is allowed (ii) repetition is not allowed? How many 3-digit even numbers can be formed from 1, 2, …, 7 with repetition? [3 marks]

Try it first, then show the solutionSolution

1. (i) $6\times 6\times 6=216$
2. (ii) $6\times 5\times 4=120$
3. Even: units digit from {2, 4, 6} (3 ways), other two digits 7 ways each: $7\times 7\times 3=147$

Answer: 216; 120; 147

Factorial notation

NCERT section 6.3.2

For a natural number n, $n!=1\times 2\times 3\times \cdots \times n$ (read 'n factorial'). By definition, $0!=1$.

• $n!=n\times (n-1)!$, so $7!=7\times 6!$, and $\frac{7!}{5!}=7\times 6=42$.
• Values: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040.
• Factorials do not add or multiply like ordinary numbers: $3!+4!=30$, but $7!=5040$.

Tip for equations with factorials: divide or multiply through by the largest factorial so that everything becomes a product of a few numbers.

Worked example

If $\frac{1}{7!}+\frac{1}{8!}=\frac{x}{9!}$, find x. [2 marks]

Try it first, then show the solutionSolution

1. Multiply both sides by 9!: $\frac{9!}{7!}+\frac{9!}{8!}=x$
2. $9\times 8+9=72+9=81$

Answer: x = 81

Permutations of distinct objects: nPr

NCERT section 6.3.1 and 6.3.3

A permutation is an arrangement in a definite order of some or all of a set of objects.

Theorem: the number of permutations of n different objects taken r at a time ($0<r\le n$), without repetition, is
${}^n{P}_r=n(n-1)(n-2)\dots (n-r+1)=\frac{n!}{(n-r)!}$

Proof idea: fill r places in order. The first place can be filled in n ways, the second in $n-1$ ways (one object is used), …, the r-th in $n-r+1$ ways. By the counting principle, multiply. Multiplying and dividing by $(n-r)!$ gives the factorial form.

• ${}^n{P}_n=n!$ (arranging all n); ${}^n{P}_0=1$.
• With repetition allowed, each of the r places has n choices: $n^r$ arrangements.

Worked example

Find n if ${}^n{P}_4=20\times {}^n{P}_2$. [2 marks]

Try it first, then show the solutionSolution

1. $n(n-1)(n-2)(n-3)=20n(n-1)\Rightarrow (n-2)(n-3)=20$ (n ≥ 4, so $n(n-1)\ne 0$)
2. $n^2-5n-14=0\Rightarrow (n-7)(n+2)=0$, so n = 7 (n cannot be negative)

Answer: n = 7

Arrangements with conditions

NCERT section 6.3

Common conditions and how to handle them:

• Fixed positions (word starts with a vowel): fill the fixed place first, then arrange the rest.
• Objects together (all vowels together): tie them into one block; arrange the block with the other objects, then multiply by the arrangements inside the block.
• Objects never together: total arrangements minus 'together' arrangements; or the gap method (arrange the others first, then place the restricted objects in the gaps).
• Particular places (vowels in even places): choose which objects go in those places, arrange them there, and arrange the rest in the remaining places.

Worked example

In how many ways can the letters of ARTICLE be arranged so that (i) the vowels are together (ii) the vowels are in the even places? [3 marks]

Try it first, then show the solutionSolution

1. ARTICLE has 7 different letters: vowels A, I, E and consonants R, T, C, L
2. (i) Treat AIE as one block: 5 units arrange in 5! ways, the vowels inside in 3! ways: $120\times 6=720$
3. (ii) Even places 2, 4, 6 take the 3 vowels in 3! ways; the 4 consonants fill the odd places in 4! ways: $6\times 24=144$

Answer: 720; 144

Permutations when objects are not all different

NCERT section 6.3.4

Theorem: the number of permutations of n objects, of which $p_1$ are alike of one kind, $p_2$ alike of a second kind, …, $p_k$ alike of a k-th kind (the rest all different), is
$\frac{n!}{p_1!p_2!\dots p_k!}$

Why: if the alike objects were labelled differently there would be n! arrangements. Swapping identical objects among themselves gives the same arrangement, and there are $p_1!$ such swaps for the first kind, $p_2!$ for the second, and so on. So each distinct arrangement has been counted $p_1!p_2!\dots$ times.

Example: BANANA has 6 letters with three A's and two N's: $\frac{6!}{3!2!}=60$ arrangements.

Worked example

Find the number of arrangements of the letters of COMMITTEE. [2 marks]

Try it first, then show the solutionSolution

1. COMMITTEE has 9 letters: M twice, T twice, E twice, and C, O, I once each
2. $\frac{9!}{2!2!2!}=\frac{362880}{8}=45360$

Answer: 45360

Combinations: nCr

NCERT section 6.4

A combination is a selection of objects where order does not matter: choosing {A, B} is the same as choosing {B, A}.

Link with permutations: each combination of r different objects can be arranged in $r!$ ways, and doing this for every combination gives every permutation exactly once. So
${}^n{P}_r={}^n{C}_r\times r!\Rightarrow {}^n{C}_r=\frac{n!}{r!(n-r)!}$

Properties:

• ${}^n{C}_0={}^n{C}_n=1$, ${}^n{C}_1=n$
• ${}^n{C}_r={}^n{C}_{n-r}$ (choosing r to take is the same as choosing $n-r$ to leave)
• If ${}^n{C}_a={}^n{C}_b$, then $a=b$ or $a+b=n$
• Pascal's rule: ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$

Worked example

If ${}^n{C}_7={}^n{C}_5$, find ${}^n{C}_2$. Verify that ${}^7{C}_3+{}^7{C}_2={}^8{C}_3$. [2 marks]

Try it first, then show the solutionSolution

1. $7\ne 5$, so $7+5=n$, n = 12; ${}^{12}{C}_2=\frac{12\times 11}{2}=66$
2. ${}^7{C}_3+{}^7{C}_2=35+21=56$ and ${}^8{C}_3=\frac{8\times 7\times 6}{6}=56$ ✓

Answer: 66; verified.

Selection problems and mixed problems

NCERT section 6.4 (Examples and Miscellaneous)

Selections from groups: select from each group separately and multiply. Choosing 3 boys from 6 and 2 girls from 5: ${}^6{C}_3\times {}^5{C}_2$.

'At least' and 'at most': split into cases (exactly 3, exactly 4, …) and add the cases.

Geometry: lines through n points with no three collinear: ${}^n{C}_2$; triangles: ${}^n{C}_3$; diagonals of an n-gon: ${}^n{C}_2-n$.

Select, then arrange: for 'words with 2 vowels and 3 consonants', first select the letters (${}^a{C}_2\times {}^b{C}_3$), then arrange the 5 chosen letters (5!).

Is it P or C? Ask: if I swap two chosen items, do I get a different outcome? Yes → permutation. No → combination.

Worked example

A committee of 5 is to be formed from 6 boys and 4 girls. In how many ways can it be done if it has (i) exactly 2 girls (ii) at least 3 girls? [3 marks]

Try it first, then show the solutionSolution

1. (i) 2 girls and 3 boys: ${}^4{C}_2\times {}^6{C}_3=6\times 20=120$
2. (ii) 3 girls and 2 boys: ${}^4{C}_3\times {}^6{C}_2=4\times 15=60$
3. 4 girls and 1 boy: ${}^4{C}_4\times {}^6{C}_1=6$; total $60+6=66$

Answer: 120; 66

Maths in real life: Permutations and Combinations

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.

Topics in this chapter

• Combinations: Selections, Committees and Grouping Problems
• Permutations: Arrangements with Conditions and Repeated Objects

Permutations and Combinations formula sheet

Fundamental principle of counting

$\text{ways for successive events}=m\times n\times p\times \dots$

Factorial

$n!=n(n-1)\dots 2\cdot 1,0!=1,n!=n(n-1)!$

Permutations

${}^n{P}_r=\frac{n!}{(n-r)!},\text{with repetition: }{n}^r$

Objects not all different

$\frac{n!}{p_1!p_2!\dots p_k!}$

Combinations

${}^n{C}_r=\frac{n!}{r!(n-r)!},{}^n{P}_r={}^n{C}_r\times r!$

Properties of nCr

${}^n{C}_r={}^n{C}_{n-r},{}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$

${}^n{C}_a={}^n{C}_b\Rightarrow a=b$ or $a+b=n$.

How the school exam asks Permutations and Combinations

The question patterns to expect, and the method that works for each.

Tricks to remember Permutations and Combinations

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

• P for Position, C for Choose

  Permutations care about Positions (order); Combinations only Choose.

  Use it when: Deciding which formula.

• AND multiplies, OR adds

  Do this AND that → multiply. Case 1 OR case 2 → add.

  Use it when: Committees with 'at least'.

• Glue and unglue

  For 'together', glue the group into one block, count, then multiply by the ways to arrange inside the glue.

  Use it when: Vowels-together questions.

• Divide out the twins

  Identical letters can't be told apart, so divide by the factorial of each repeat count.

  Use it when: MISSISSIPPI-type words.

• Take r = leave n − r

  ${}^{20}{C}_{18}={}^{20}{C}_2=190$: always compute the smaller one.

  Use it when: Quick evaluation.

Permutations and Combinations: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Permutations and Combinations

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

• ✗Mistake: Using P when order does not matter

  Not asking whether swapping changes the outcome.

  Wrong working

  Committee of 3 from 10: ${}^{10}{P}_3=720$

  Right working

  ${}^{10}{C}_3=120$

  ✓Do this: Committees, teams, selections → C; arrangements, words, numbers → P.

  Usually costs: Whole question.

• ✗Mistake: Allowing 0 as the first digit

  Treating all places the same.

  ✓Do this: Fill the first place first with non-zero digits.

  Usually costs: 1-2 marks.

• ✗Mistake: Multiplying cases instead of adding them

  Confusing 'and' with 'or'.

  ✓Do this: 'At least 3 girls' = (exactly 3) + (exactly 4): add.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting to arrange inside the block

  Counting the block as one object and stopping.

  Wrong working

  Vowels of ARTICLE together: 5! = 120

  Right working

  5! × 3! = 720

  ✓Do this: Multiply by the arrangements inside the block.

  Usually costs: 1 mark.

• ✗Mistake: Not dividing for repeated letters

  Treating identical letters as different.

  ✓Do this: Count each letter's frequency before writing the formula.

  Usually costs: 1 mark.

• ✗Mistake: Keeping an impossible root

  Not checking n against the problem.

  ✓Do this: n must be a natural number at least r; reject negative or fractional values.

  Usually costs: 1 mark.

NCERT exercise guide: Permutations and Combinations

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Permutations and Combinations

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

1. Q1 · 1 mark · MCQExam level

   The value of ${}^{10}{C}_8$ is:
   (a) 90
   (b) 45
   (c) 80
   (d) 1814400

   Hint

   Use ${}^{10}{C}_2$.

   Answer

   (b) 45

2. Q2 · 1 mark · MCQExam level

   The number of arrangements of the letters of the word DELHI that begin with D is:
   (a) 120
   (b) 24
   (c) 60
   (d) 4

   Hint

   Arrange the other four.

   Answer

   (b) 24

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): The number of straight lines through 15 points, no three collinear, is 105.
   Reason (R): Each line is fixed by choosing 2 of the points, and order does not matter.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   ${}^{15}{C}_2$.

   Answer

   (a)

4. Q4 · 2 marks · Very short answerExam level

   Using 6 flags of different colours, how many signals can be made if a signal uses (i) exactly 2 flags one below the other (ii) any number of flags, at least one?

   Hint

   Order matters.

   Answer

   (i) ${}^6{P}_2=30$ (ii) $6+30+120+360+720+720=1956$

5. Q5 · 2 marks · Very short answerHarder exam level

   Find r if ${}^6{P}_r=2\times {}^5{P}_r$.

   Hint

   Write in factorials.

   Answer

   $\frac{6!}{(6-r)!}=\frac{2\cdot 5!}{(5-r)!}\Rightarrow \frac{6}{6-r}=2\Rightarrow r=3$. Check: ${}^6{P}_3=120=2\times 60=2\times {}^5{P}_3$.

6. Q6 · 2 marks · Very short answerHarder exam level

   In how many ways can 4 boys and 2 girls be selected from 7 boys and 5 girls?

   Hint

   Multiply.

   Answer

   ${}^7{C}_4\times {}^5{C}_2=35\times 10=350$.

7. Q7 · 2 marks · Very short answerHarder exam level

   How many words can be formed from the letters of FRACTION, using each letter once, so that the vowels are together and the consonants are together?

   Hint

   Two blocks.

   Answer

   Vowels A, I, O (block, 3!); consonants F, R, C, T, N (block, 5!); the two blocks in 2! orders: $2\times 6\times 120=1440$.

8. Q8 · 3 marks · Short answerHarder exam level

   In how many ways can the letters of EXAMINATION be arranged if the words start with E and end with N?

   Hint

   A and I appear twice among the middle letters.

   Answer

   Fix E first and one N last; the middle 9 letters X, A, A, M, I, I, N, T, O have A twice and I twice: $\frac{9!}{2!2!}=90720$.

9. Q9 · 3 marks · Short answerHarder exam level

   A bag has 6 black and 7 red balls. In how many ways can 2 black and 3 red balls be selected? How many of these selections include a particular red ball?

   Hint

   Fix the particular ball.

   Answer

   ${}^6{C}_2\times {}^7{C}_3=15\times 35=525$. With a particular red ball: ${}^6{C}_2\times {}^6{C}_2=15\times 15=225$.

10. Q10 · 3 marks · Short answerHOTS / competency

    How many triangles can be formed from 10 points of which exactly 4 lie on one straight line (no other three collinear)?

    Hint

    Subtract the 'triangles' made from collinear points.

    Answer

    ${}^{10}{C}_3-{}^4{C}_3=120-4=116$.

11. Q11 · 2 marks · Very short answerHOTS / competency

    If the letters A, C, T are arranged in all possible ways and the words listed as in a dictionary, what is the rank of CAT? Generalise: how would you find the rank of a word with distinct letters?

    Hint

    Count words before it.

    Answer

    Words starting with A: 2! = 2 (ACT, ATC). Then CAT comes first among words starting with C, so its rank is 3. In general, for each position count the arrangements of the remaining letters that start with a smaller letter, add them all, then add 1.

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. 1.What is 0!?Show answer

   1.

2. 2.Write ${}^n{P}_r$ in factorials.Show answer

   $\frac{n!}{(n-r)!}$.

3. 3.How are ${}^n{P}_r$ and ${}^n{C}_r$ related?Show answer

   ${}^n{P}_r={}^n{C}_r\times r!$.

4. 4.Arrangements of the letters of BANANA?Show answer

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

5. 5.Value of ${}^{15}{C}_{13}$?Show answer

   ${}^{15}{C}_2=105$.

6. 6.Diagonals in an octagon?Show answer

   ${}^8{C}_2-8=20$.

7. 7.State Pascal's rule.Show answer

   ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$.

8. 8.Number of 3-letter arrangements of 5 letters with repetition?Show answer

   $5^3=125$.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state and use the fundamental principle of counting?
• ☐Can you simplify expressions like $\frac{8!}{6!2!}$?
• ☐Can you derive ${}^n{P}_r=\frac{n!}{(n-r)!}$?
• ☐Can you count numbers with restrictions on digits?
• ☐Can you count words with letters together or not together?
• ☐Can you count arrangements with repeated letters?
• ☐Can you derive ${}^n{C}_r$ from ${}^n{P}_r$?
• ☐Can you solve ${}^n{C}_a={}^n{C}_b$ and ratio equations?
• ☐Can you count committees with 'at least' conditions?
• ☐Can you decide whether a question needs P or C?

Permutations and Combinations: frequently asked questions