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CBSE · Class 11 · Mathematics

Permutations (Ch 6: 6.3)

4 concepts · 32 practice questions with answers · Free to read

Permutations of distinct objects

From daily life

Choosing a president and a secretary from 5 students: the order matters because the posts are different. Each different order is a permutation.

Simple definition

A permutation is an arrangement of objects in a definite order. The number of permutations of n distinct objects taken r at a time is ⁿPᵣ = n!/(n − r)!. When r = n it is n!.

Formula: ⁿPᵣ = n! / (n − r)!

Rule 1: Compute ⁿPᵣ — Multiply r numbers starting from n and going down.

  • Easy: ⁵P₂ = 5 × 4 = 20.
  • Medium: ⁷P₃ = 7 × 6 × 5 = 210.
  • Exam-type: ¹⁰P₄ = 10 × 9 × 8 × 7 = 5040.

Rule 2: Arrangement problems — Decide n and r, then use ⁿPᵣ.

  • Easy: Ways to give 1st, 2nd and 3rd prize among 8 students: ⁸P₃ = 336.
  • Medium: 3-letter words (no repeat) from 5 letters A, B, C, D, E: ⁵P₃ = 60.
  • Exam-type: Arranging all 6 books on a shelf: 6! = 720.
Remember

ⁿPᵣ = n × (n − 1) × … (r factors). Also ⁿP₁ = n and ⁿPₙ = n!.

Common mistake

Using permutations when order does not matter.

Memory clue

P is for Position: the order matters.

Practice questions

Easy
  1. 1Find ⁶P₂.
  2. 2Find ⁸P₃.
  3. 3In how many ways can 4 people stand in a line?
Medium
  1. 4How many 4-digit numbers can be made from 1, 2, 3, 4, 5 without repetition?
  2. 5If ⁿP₂ = 42, find n.
  3. 6How many 3-letter code words (no repeat) can be made from the letters of the word PLANET?
Challenging
  1. 7In how many ways can a president, vice-president and treasurer be chosen from 9 members?
  2. 8If ⁿP₃ = 6 × ⁿ⁻¹P₂, find n.
Show answers (try them yourself first)
  1. 130
  2. 2336
  3. 324
  4. 4120
  5. 5n(n − 1) = 42, so n = 7
  6. 6120
  7. 7504
  8. 8n(n − 1)(n − 2) = 6(n − 1)(n − 2), so n = 6

Arrangements with repetition allowed

From daily life

A phone PIN can use the same digit again and again. Each place has all the choices again.

Simple definition

If each of r places can be filled in n ways independently (repeats allowed), the total is nʳ.

Formula: nʳ

Rule 1: Use nʳ — Count choices per place, then multiply.

  • Easy: 4-digit PINs from 0–9: 10⁴ = 10000.
  • Medium: 3-letter strings from A, B, C, D: 4³ = 64.
  • Exam-type: Ways to answer 5 four-option questions: 4⁵ = 1024.

Rule 2: Mixed conditions — Fix the restricted place first.

  • Easy: 3-digit numbers (digits 0–9, repeat allowed): 9 × 10 × 10 = 900.
  • Medium: Even 3-digit numbers: 9 × 10 × 5 = 450.
  • Exam-type: 3-digit numbers with digits from {1, 2, 3} that are less than 300: 2 × 3 × 3 = 18.
Remember

If repeats are not allowed, use ⁿPᵣ; if they are, use nʳ.

Common mistake

Using nʳ when the same object cannot be used again.

Memory clue

Repeat? Power. No repeat? P.

Practice questions

Easy
  1. 1How many 2-digit PINs are possible with digits 1 to 5 (repeat allowed)?
  2. 2How many 4-letter strings can be made from A, B, C?
  3. 3In how many ways can 3 letters be posted in 4 letter boxes?
Medium
  1. 4How many 5-digit numbers exist?
  2. 5How many 3-digit numbers can be made from 0, 1, 2, 3 with repetition?
  3. 6How many 3-digit odd numbers use digits 1 to 9 with repetition?
Challenging
  1. 7In how many ways can 6 children each choose one of 3 clubs?
  2. 8How many 4-digit numbers have all digits even (first digit non-zero)?
Show answers (try them yourself first)
  1. 125
  2. 281
  3. 364
  4. 490000
  5. 548
  6. 6405
  7. 7729
  8. 8500

Arrangements with identical objects

From daily life

The letters of "BOOK": swapping the two Os gives the same word. So we divide by the ways the identical letters can swap.

Simple definition

If among n objects, p are alike of one kind, q alike of another and r alike of a third, the number of arrangements is n!/(p! q! r!).

Formula: n! / (p! q! r!)

Rule 1: Words from letters — Count the total letters and the repeats.

  • Easy: BOOK: 4!/2! = 12.
  • Medium: BANANA: 6!/(3! 2!) = 60.
  • Exam-type: MISSISSIPPI: 11!/(4! 4! 2!) = 34650.

Rule 2: Paths and objects — Use the same idea for coloured balls and grid paths.

  • Easy: 3 red and 2 blue balls in a row: 5!/(3! 2!) = 10.
  • Medium: Grid path 3 right and 2 up: 5!/(3! 2!) = 10.
  • Exam-type: Arrange 1, 1, 2, 2, 3: 5!/(2! 2!) = 30.
Remember

Always divide by the factorials of the repeated counts.

Common mistake

Forgetting to divide for repeated letters.

Memory clue

Same twins? Divide by the swap count.

Practice questions

Easy
  1. 1Arrangements of the letters of LEVEL.
  2. 2Arrangements of the letters of APPLE.
  3. 3Arrangements of the letters of BALLOON.
Medium
  1. 4How many 6-digit numbers use 1, 1, 2, 2, 3, 3 each exactly the given number of times?
  2. 5Arrangements of 4 red and 3 green flags in a line.
  3. 6Arrangements of the letters of INDIA.
Challenging
  1. 7How many arrangements of the letters of MATHEMATICS?
  2. 8How many arrangements of the letters of ARRANGE?
Show answers (try them yourself first)
  1. 15!/(2! 2!) = 30
  2. 25!/2! = 60
  3. 37!/(2! 2!) = 1260
  4. 46!/(2! 2! 2!) = 90
  5. 57!/(4! 3!) = 35
  6. 65!/2! = 60
  7. 711!/(2! 2! 2!) = 4989600
  8. 87!/(2! 2!) = 1260

Arrangements with conditions

From daily life

Seat 5 friends in a row, but two best friends must sit together. Treat the pair as one block, and then swap them inside the block.

Simple definition

Together: glue the objects into one block, arrange the blocks, then arrange inside the block. Not together: total − together. A fixed position: fill it first.

Rule 1: Together — Block method.

  • Easy: 5 people with A and B together: 4! × 2! = 48.
  • Medium: Letters of MOBILE with all vowels together: 4! × 3! = 144.
  • Exam-type: 4 boys and 3 girls with all girls together: 5! × 3! = 720.

Rule 2: Not together or fixed place — Subtract or fix.

  • Easy: 5 people with A and B not together: 120 − 48 = 72.
  • Medium: 6 books with a given book first: 5! = 120.
  • Exam-type: Word of 5 letters ABCDE starting with A and ending with E: 3! = 6.
Remember

Count the block first, then the arrangement inside it.

Common mistake

Forgetting the inside arrangement of the block.

Memory clue

Glue, arrange, shake.

Practice questions

Easy
  1. 14 people with 2 specific people together. How many arrangements?
  2. 26 people in a row, 2 specific people never together.
  3. 3How many arrangements of the letters of DELHI with vowels at the ends? (E and I at ends)
Medium
  1. 4How many arrangements of the letters of TRIANGLE with the vowels together?
  2. 55 boys and 4 girls in a row, no two girls together.
  3. 6How many 4-digit numbers from 1 to 5 (no repeat) are even?
Challenging
  1. 7In how many ways can 3 men and 3 women sit alternately in a row (any start)?
  2. 8How many arrangements of the letters of MOTHER have T and H together?
Show answers (try them yourself first)
  1. 112
  2. 2480
  3. 32! × 3! = 12
  4. 46! × 3! = 4320
  5. 55! × ⁶P₄ = 43200
  6. 62 × ⁴P₃ = 48
  7. 72 × 3! × 3! = 72
  8. 85! × 2! = 240