Permutations and Combinations ISC Class 11: Counting Made Clear

What Permutations and Combinations is about

Weightage: Permutations and Combinations belongs to the unit Algebra, which CISCE weights at 26 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Fundamental principle of counting

Multiplication principle: if an event can occur in m ways and, following it, a second event can occur in n ways, the two events in succession can occur in m × n ways; this extends to any number of stages. Addition principle: if two jobs are mutually exclusive alternatives (you do one or the other), the number of ways is m + n. Use 'and' → multiply, 'or' → add. Always ask whether repetition is allowed, because that changes the number of choices at later stages.

Worked example

A 3-digit lock code uses the digits 0 to 9. How many codes are possible (i) with repetition (ii) without repetition?

Try it first, then show the solutionSolution

1. (i) Each of three places has 10 choices: 10 × 10 × 10 = 1000
2. (ii) 10 choices, then 9, then 8: 10 × 9 × 8 = 720

Answer: 1000; 720

Factorial notation

n! = n × (n − 1) × … × 2 × 1 for a natural number n, and 0! = 1 by definition (so that formulae such as ⁿPₙ = n!/0! work). The key identity n! = n × (n − 1)! lets you cancel factorials quickly: write the larger factorial down to the smaller one and cancel. Factorials grow very fast: 10! = 3 628 800.

Worked example

Find n if (n + 1)! = 12 × (n − 1)!.

Try it first, then show the solutionSolution

1. (n + 1)! = (n + 1) n (n − 1)!
2. (n + 1) n = 12 ⇒ n² + n − 12 = 0 ⇒ (n + 4)(n − 3) = 0
3. n is natural, so n = 3

Answer: n = 3

Permutations ⁿPᵣ and number formation

A permutation is an arrangement in a definite order. Filling r places from n different objects without repetition: the first place has n choices, the second n − 1, …, the r-th n − r + 1, so ⁿPᵣ = n(n − 1)…(n − r + 1) = n!/(n − r)!, 0 ≤ r ≤ n. With repetition allowed the count is nʳ. When forming numbers, fill the restricted place first: the leading digit cannot be 0, an even number must end in an even digit, a number divisible by 5 must end in 0 or 5.

Worked example

How many 4-digit numbers with distinct digits can be formed from 1, 2, 3, 4, 5, 6, 7? How many of them are even?

Try it first, then show the solutionSolution

1. All: ⁷P₄ = 7 × 6 × 5 × 4 = 840
2. Even: units digit from {2, 4, 6} (3 ways); remaining three places from the other 6 digits: ⁶P₃ = 120
3. 3 × 120 = 360

Answer: 840; 360

Restricted arrangements: together and never together

Always together: tie the k objects into one block, arrange the block with the others, then multiply by k! for the order inside the block. Never together (as a group): total arrangements minus 'all together'. No two together: arrange the other objects first, then place the restricted ones in the gaps between and at the ends: m objects create m + 1 gaps, so the count is m! × ᵐ⁺¹Pₖ. Read the wording carefully: 'never all together' and 'no two together' are different conditions.

Worked example

5 boys and 3 girls stand in a row. In how many ways can this be done if (i) the girls are together (ii) no two girls are together?

Try it first, then show the solutionSolution

1. (i) Block of 3 girls + 5 boys = 6 units: 6! × 3! = 720 × 6 = 4320
2. (ii) Arrange boys: 5! = 120, creating 6 gaps; place 3 girls in 3 of the gaps: ⁶P₃ = 120
3. 120 × 120 = 14 400

Answer: 4320; 14 400

Permutations of alike and repeated things

If n objects contain p alike of one kind, q alike of another and r alike of a third (the rest different), the number of distinct arrangements of all n is n!/(p! q! r!): swapping identical objects gives nothing new, so we divide out those swaps. This is how words with repeated letters are counted. For conditions such as 'two particular letters together', glue them and recount the repeats in the new list.

Worked example

How many arrangements of the letters of BANANA are there? In how many are the two N's together?

Try it first, then show the solutionSolution

1. B ×1, A ×3, N ×2: 6!/(3! 2!) = 720/12 = 60
2. Glue NN: units B, A, A, A, (NN): 5!/3! = 20

Answer: 60; 20

Circular permutations

In a circle there is no first place: rotating everyone one seat gives the same arrangement. Fix one person and arrange the rest: (n − 1)! circular arrangements of n different objects when clockwise and anticlockwise orders are distinguishable (people at a round table). When they are not distinguishable (beads on a necklace, flowers on a garland, which can be turned over), the count is (n − 1)!/2. For 'two particular people together' at a round table, glue them: (n − 2)! × 2.

Worked example

(i) In how many ways can 6 people sit at a round table? (ii) In how many of these do two particular people sit together? (iii) How many necklaces can be made from 7 different beads?

Try it first, then show the solutionSolution

1. (i) (6 − 1)! = 120
2. (ii) Glue the pair: 5 units round the table: 4! × 2 = 48
3. (iii) (7 − 1)!/2 = 720/2 = 360

Answer: 120; 48; 360

Combinations ⁿCᵣ and their properties

A combination is a selection where order does not matter. Each selection of r objects can be arranged in r! ways, so ⁿPᵣ = r! × ⁿCᵣ, giving ⁿCᵣ = n!/(r!(n − r)!). Properties: ⁿC₀ = ⁿCₙ = 1; ⁿCᵣ = ⁿCₙ₋ᵣ (choosing r to take is choosing n − r to leave); ⁿCₓ = ⁿCᵧ ⇒ x = y or x + y = n; Pascal's rule ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ⁺¹Cᵣ. Use ⁿCᵣ = ⁿCₙ₋ᵣ to shorten work: ²⁰C₁₈ = ²⁰C₂ = 190.

Worked example

If ⁿC₉ = ⁿC₇, find n and ⁿC₂.

Try it first, then show the solutionSolution

1. 9 ≠ 7, so 9 + 7 = n, n = 16
2. ¹⁶C₂ = 16 × 15/2 = 120

Answer: n = 16; 120

Selections with restrictions, identical objects and mixed problems

At least / at most: split into cases and add, or use total minus the unwanted cases. Never 'choose one first and then choose the rest freely', which counts some selections more than once. Things not all different: from p alike, q alike (of another kind) and r different objects, the number of selections of any size, including none, is (p + 1)(q + 1)2ʳ; subtract 1 to exclude the empty selection. Mixed problems: select with C, then arrange with ! or P.

Worked example

How many 5-letter words, each with 2 different vowels and 3 different consonants, can be formed from 4 vowels and 6 consonants?

Try it first, then show the solutionSolution

1. Select vowels: ⁴C₂ = 6; select consonants: ⁶C₃ = 20
2. Arrange the 5 chosen letters: 5! = 120
3. 6 × 20 × 120 = 14 400

Answer: 14 400

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 and Mixed Problems
• Permutations: Arrangements and Restricted Cases

Permutations and Combinations formula sheet

Counting principle

$\text{and}\Rightarrow m\times n;\text{or}\Rightarrow m+n$

Stages multiply, alternatives add.

Factorial

$n!=n(n-1)!,0!=1$

Cancel the larger factorial down to the smaller.

Permutations

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

With repetition: nʳ.

Alike objects

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

Divide by the factorial of each repeat count.

Circular arrangements

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

Halve only if the ring can be turned over.

Combinations

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

ⁿCᵣ = ⁿCₙ₋ᵣ.

Pascal's rule

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

Builds Pascal's triangle.

Selections from alike things

$(p+1)(q+1)2^r-1$

Non-empty selections from p alike, q alike, r different.

Permutations and Combinations: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

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

  If swapping two chosen items gives a different outcome, use P; if not, use C.

  Use it when: Deciding between permutation and combination.

• Glue and jiggle

  Glue the 'together' items into one block, then jiggle them inside it (k!).

  Use it when: Always-together problems.

• Gaps for gaps

  For 'no two together', seat the others first and drop the restricted ones into the gaps.

  Use it when: Never-adjacent problems.

• Circle: fix one

  Round table: fix one person, then (n − 1)!; flip-able ring: halve it.

  Use it when: Circular permutations.

• Total minus bad

  'At least one' = all − none.

  Use it when: At-least selections.

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 D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 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 where C is needed (or vice versa)

  Students do not ask whether order matters.

  Wrong working

  Choosing 3 prefects from 10: ¹⁰P₃ = 720.

  Right working

  Prefects are a selection: ¹⁰C₃ = 120.

  ✓Do this: Swap two chosen items: if nothing changes, use C.

  Usually costs: Full marks for the part.

• ✗Mistake: Overcounting 'at least one'

  Choosing one woman first and then any others counts the same committee several times.

  Wrong working

  At least one girl in a team of 4 from 7 boys, 5 girls: 5 × ¹¹C₃ = 825.

  Right working

  ¹²C₄ − ⁷C₄ = 495 − 35 = 460.

  ✓Do this: Use cases or total minus complement.

  Usually costs: 2-3 marks.

• ✗Mistake: Letting 0 lead a number

  Students use ⁿPᵣ directly when 0 is among the digits.

  Wrong working

  3-digit numbers from 0 to 5, distinct: ⁶P₃ = 120.

  Right working

  5 × 5 × 4 = 100.

  ✓Do this: Fill the leading place first.

  Usually costs: 1-2 marks.

• ✗Mistake: Using n! for a round table

  Rotations of the same seating are counted separately.

  Wrong working

  6 people round a table: 720.

  Right working

  (6 − 1)! = 120.

  ✓Do this: Fix one person first.

  Usually costs: 1 mark.

• ✗Mistake: Halving for a round table of people

  Students apply the necklace rule where clockwise and anticlockwise are different.

  Wrong working

  6 people round a table: 5!/2 = 60.

  Right working

  120: a person's left and right neighbours are different arrangements.

  ✓Do this: Halve only for objects that can be turned over.

  Usually costs: 1 mark.

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 0! is:
   (a) 0
   (b) 1
   (c) undefined
   (d) ∞

   Hint

   Definition.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   ¹⁰C₈ equals:
   (a) 45
   (b) 90
   (c) 80
   (d) 1

   Hint

   ¹⁰C₂.

   Answer

   (a)

3. Q3 · 1 mark · Very short answerExam level

   Evaluate 8!/(6! × 2!).

   Hint

   Cancel 6!.

   Answer

   8 × 7/2 = 28

4. Q4 · 1 mark · Very short answerExam level

   In how many ways can 5 people sit at a round table?

   Hint

   (n − 1)!.

   Answer

   4! = 24

5. Q5 · 2 marks · Short answerExam level

   How many 4-letter arrangements (without repetition) can be made from the letters of MONDAY? How many use all six letters?

   Hint

   All letters are different.

   Answer

   ⁶P₄ = 360; 6! = 720

6. Q6 · 2 marks · Short answerHarder exam level

   If ⁿC₁₀ = ⁿC₁₂, find ²³Cₙ.

   Hint

   x + y = n.

   Answer

   n = 22; ²³C₂₂ = ²³C₁ = 23

7. Q7 · 2 marks · Short answerHarder exam level

   How many diagonals does a polygon with 10 sides have?

   Hint

   Lines joining vertices minus sides.

   Answer

   ¹⁰C₂ − 10 = 45 − 10 = 35

8. Q8 · 3 marks · Short answerHarder exam level

   How many 4-digit numbers with distinct digits, formed from 1, 2, 3, 4, 5, 6, lie between 4000 and 5000?

   Hint

   The first digit must be 4.

   Answer

   First digit 4 (1 way); the other three from the remaining 5 digits: ⁵P₃ = 60

9. Q9 · 3 marks · Short answerHarder exam level

   Find the number of arrangements of the letters of ARRANGE. In how many are the two R's not together?

   Hint

   A ×2, R ×2.

   Answer

   Total 7!/(2! 2!) = 1260. R's together: 6!/2! = 360. Not together: 1260 − 360 = 900.

10. Q10 · 2 marks · Short answerHarder exam level

    In how many ways can (i) 7 people sit at a round table (ii) 7 different beads be strung into a necklace?

    Hint

    Halve only for the necklace.

    Answer

    (i) 6! = 720 (ii) 6!/2 = 360

11. Q11 · 3 marks · Short answerHarder exam level

    A basket has 3 identical apples, 4 identical oranges and 2 different mangoes. In how many ways can at least one fruit be selected?

    Hint

    (p + 1)(q + 1)2ʳ − 1.

    Answer

    4 × 5 × 2² − 1 = 79

12. Q12 · 5 marks · Long answerHarder exam level

    A box holds 5 red, 4 blue and 3 green marbles, all numbered and so all different.
    (a) In how many ways can 3 marbles of three different colours be chosen? [1]
    (b) In how many ways can 4 marbles be chosen with exactly 2 red ones? [2]
    (c) In how many ways can 4 marbles be chosen with at least one of each colour? [2]

    Hint

    In (c), one colour appears twice: split into three cases by which colour it is.

    Answer

    (a) 5 × 4 × 3 = 60 (b) ⁵C₂ × ⁷C₂ = 10 × 21 = 210 (c) two red: ⁵C₂ × 4 × 3 = 120; two blue: 5 × ⁴C₂ × 3 = 90; two green: 5 × 4 × ³C₂ = 60; total 120 + 90 + 60 = 270

13. Q13 · 3 marks · Short answerHarder exam level

    In how many ways can the letters of EXAMINATION be arranged so that the vowels occupy only the odd positions?

    Hint

    6 vowels (A, I repeated) and 6 odd positions.

    Answer

    Vowels E, A, A, I, I, O fill positions 1, 3, …, 11: 6!/(2! 2!) = 180. Consonants X, M, N, N, T fill the 5 even positions: 5!/2! = 60. Total 180 × 60 = 10 800.

14. Q14 · 3 marks · Short answerHOTS / competency

    12 points lie in a plane, and exactly 5 of them are collinear. How many triangles can be formed with vertices at these points?

    Hint

    Remove the collinear triples.

    Answer

    ¹²C₃ − ⁵C₃ = 220 − 10 = 210

15. Q15 · 3 marks · Short answerHOTS / competency

    In how many ways can 6 boys and 4 girls stand in a row so that no two girls are next to each other?

    Hint

    Seat the boys, then use the 7 gaps.

    Answer

    6! × ⁷P₄ = 720 × 840 = 604 800

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.ⁿPᵣ in factorials?Show answer

   n!/(n − r)!.

2. 2.ⁿCᵣ in factorials?Show answer

   n!/(r!(n − r)!).

3. 3.Link between P and C?Show answer

   ⁿPᵣ = r! × ⁿCᵣ.

4. 4.Arrangements of n objects with p alike?Show answer

   n!/p!.

5. 5.n people round a table?Show answer

   (n − 1)!.

6. 6.Necklace of n different beads?Show answer

   (n − 1)!/2.

7. 7.Pascal's rule?Show answer

   ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ⁺¹Cᵣ.

8. 8.If ⁿCₓ = ⁿCᵧ?Show answer

   x = y or x + y = n.

Quick revision checklist

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

• ☐Can you decide whether a problem needs P or C?
• ☐Can you solve for n in an equation involving factorials?
• ☐Can you count numbers formed from digits including 0?
• ☐Can you count arrangements with objects together, and with no two together?
• ☐Can you count words with repeated letters?
• ☐Can you distinguish a round table from a necklace?
• ☐Can you prove Pascal's rule?
• ☐Can you count 'at least' selections without overcounting?

Permutations and Combinations: frequently asked questions