Find ⁵C₃.

10

⁵C₃ = 5! / (3! × 2!) = (5 × 4)/(2 × 1) = 10.

Understanding Combinatorics questions

Combinatorics questions ask how many arrangements or selections are possible. The whole decision comes down to one question: does order matter?

How to approach it

1. Decide whether order matters — permutation if yes, combination if no.
2. Apply nPr = n!/(n−r)! or nCr = n!/(r!(n−r)!).
3. Handle restrictions first, then arrange the remaining items.
4. Use the complement rule when "at least one" appears in the question.

Common mistakes to avoid

• Using a permutation where a combination is required.
• Double counting arrangements of identical items.
• Forgetting that 0! equals 1.

Where it appears: Standard in CBSE Class 11, IB HL and competitive aptitude tests.