Permutations and Combinations
Permutations and Combinations tests counting without listing: arrangements with restrictions, selections, circular arrangements, and arrangements of letters with repetition.
Core formulas
nPr = n!/(n − r)! counts ordered selections; nCr = n!/(r!(n − r)!) counts unordered ones. Key identities: nCr = nC(n − r), nCr + nC(r − 1) = (n+1)Cr, and Σ nCr = 2ⁿ. Circular permutations of n distinct objects give (n − 1)!; for a necklace or garland, where clockwise and anticlockwise orders coincide, it is (n − 1)!/2. Arrangements of n letters with p and q identical copies give n!/(p!q!).
Strategy
Fix the restriction first - seat a particular person, or choose the compulsory members - then arrange the remainder. For 'at least one' or 'at most', subtract the complement from the total or add the separate valid cases.
Worked example
A committee of 4 is chosen from 5 men and 4 women with at least 2 women. Case 2 women + 2 men: 4C2 × 5C2 = 6 × 10 = 60. Case 3 women + 1 man: 4C3 × 5C1 = 4 × 5 = 20. Case 4 women: 4C4 = 1. Total = 60 + 20 + 1 = 81 committees.
Traps
Using nPr when order is irrelevant; double counting when a selection is split into equal-sized groups (divide by k! for k equal groups); treating repeated letters as distinct; forgetting that a necklace and its reversal are the same arrangement.