Permutations and Combinations for CA Foundation
Counting arrangements (permutations) and selections (combinations), including repeated objects, circular arrangements and restricted selections.
Quantitative AptitudePaper 3, Business Mathematics12 MCQs2 lessons2 flashcards3 memory tricks
Practise Permutations and Combinations free
What you'll learn
- Permutations
Count arrangements where order matters. - Combinations
Count selections where order does not matter.
Key formulas
| Factorial | n! = n × (n − 1) × … × 1; 0! = 1 |
|---|---|
| Permutations | ⁿPᵣ = n!/(n − r)! |
| With repetition allowed | nʳ |
| Identical objects | n!/(p! q! …) |
| Circular | (n − 1)!; necklace or garland (n − 1)!/2 |
| Combinations | ⁿCᵣ = n!/(r!(n − r)!) |
| Complement | ⁿCᵣ = ⁿCₙ₋ᵣ |
| Pascal's rule | ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ |
Concepts explained
Permutations: arrangements
A permutation is an arrangement where order matters: ⁿPᵣ = n!/(n − r)!.
What it means
- Fundamental principle of counting: m ways and then n ways → m × n.
- No repetition: ⁿPᵣ; repetition allowed: nʳ.
- Identical objects: n!/(p! q! …).
- 0! = 1, so ⁿPₙ = n!.
- 3-digit numbers from 1–5 without repetition
Answer: ⁵P₃ = 60
- Arrangements of BANANA
Answer: 6!/(3! 2!) = 60
Related terms
- Factorial: product of all whole numbers up to n
- Restricted arrangements: fix the restricted positions first
Exam tip: If swapping two items gives a different result, it's a permutation.
Combinations: selections
A combination is a selection where order doesn't matter: ⁿCᵣ = n!/(r!(n − r)!).
What it means
- ⁿCᵣ = ⁿCₙ₋ᵣ; if ⁿCₐ = ⁿC_b with a ≠ b, then n = a + b.
- ⁿPᵣ = r! × ⁿCᵣ.
- Lines from n points: ⁿC₂; triangles: ⁿC₃; diagonals: n(n − 3)/2; handshakes: ⁿC₂.
- 'At least one' = total − none.
- 3 from 5 men and 4 women with exactly 2 men
- ⁵C₂ × ⁴C₁
Answer: 40
- Hexagon
- 6 × 3/2
Answer: 9
Related terms
- Pascal's rule: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ
- Binomial coefficients: ⁿCᵣ in (a + b)ⁿ
Exam tip: Committees, teams and groups are combinations; queues, numbers and words are permutations.
Circular arrangements
Arranging n distinct objects in a circle gives (n − 1)! ways, because rotating everyone doesn't create a new arrangement.
What it means
- Fix one object, arrange the remaining n − 1: (n − 1)!.
- If clockwise and anticlockwise are the same (necklace, garland): (n − 1)!/2.
- Two particular people together: treat them as one unit, then multiply by 2!.
- 6 people
Answer: 5! = 120
- 7 different beads
Answer: 6!/2 = 360
Related terms
- Linear arrangement: n! for a straight line
- Rotation: same arrangement when everyone shifts one seat
Exam tip: Ask whether flipping the circle over gives the same arrangement: if yes, divide by 2.
Memory tricks
Arrangements → ⁿPᵣ; selections → ⁿCᵣ.
Circular arrangements = (n − 1)!.
Arrangements with repeats = n!/(p! q! …).
Try 3 CA Foundation-style questions
Q1. ⁿPᵣ ÷ ⁿCᵣ is equal to
- n!
- r!
- (n − r)!
- 1
Show answer
Answer: (2) r!
ⁿPᵣ = r! × ⁿCᵣ: every selection of r things can be arranged in r! ways.
Q2. A committee of 3 is to be chosen from 5 men and 4 women with exactly 2 men. The number of ways is
- 40
- 84
- 10
- 20
Show answer
Answer: (1) 40
Choose 2 of 5 men (⁵C₂ = 10) and 1 of 4 women (⁴C₁ = 4): 10 × 4 = 40.
Q3. The number of ways to arrange the letters of the word BANANA is
- 720
- 60
- 120
- 360
Show answer
Answer: (2) 60
6 letters with A repeated 3 times and N twice: 6!/(3! × 2!) = 720/12 = 60.
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