nCr
Permutations & Combinations
Arrangements, selections, binomial theorem, and Pascal's triangle
Key Concepts
- Permutations count ordered arrangements; combinations count unordered selections
- Circular arrangements fix one position to remove rotational symmetry
- Stars and bars counts non-negative integer solutions
- Binomial coefficients satisfy Pascal's rule
Important Formulae
| Permutations / combinations | ⁿPᵣ = n!/(n−r)!; ⁿCᵣ = n!/(r!(n−r)!) |
| Circular arrangements | (n − 1)! for n distinct objects |
| Stars and bars | x₁+…+xₖ = n ⇒ C(n+k−1, k−1) solutions |
Quick Tips
- ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ — choose the smaller of r and n−r to compute faster.
- Decide first whether order matters: that picks permutation vs combination.
Sample Practice Questions
-
In the expansion (a + b + c)^3, how many distinct terms are there?
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Answer: 10
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P(5, 3) = ?
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Answer: 60
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The sum of entries in the nth row of Pascal's triangle is 2^n. What is the sum for n=7?
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Answer: 128
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The number of permutations of AABBBCC is:
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Answer: 7!/(2!3!2!) = 210
Practice Questions
Practise PRMO questions on Permutations & Combinations. Answers are revealed after each question.
Start Practice →Combinatorics