Permutations and Combinations
Permutations and combinations count arrangements and selections without listing them. The exam tests the fundamental principle of counting, factorial notation, restricted arrangements, and the decision whether order matters - a distinction that decides between nPr and nCr.
Core rules
- Fundamental principle: if one task can be done in m ways and a second in n ways, the two together can be done in m × n ways.
- Permutations (order matters): nPr = n!/(n − r)!. Combinations (order does not matter): nCr = n!/(r!(n − r)!).
- nC0 = nCn = 1 and 0! = 1. Also nCr = nC(n−r).
- Circular arrangements of n distinct objects: (n − 1)!. Arrangements of n objects with p alike of one kind and q alike of another: n!/(p!q!).
- “At least one” is usually handled by the complement: total − none.
Worked example
A firm has 8 candidates for an audit committee. (i) Selecting 3 members: order is irrelevant, so the number is 8C3 = 8!/(3!5!) = (8 × 7 × 6)/(3 × 2 × 1) = 56. (ii) Assigning 3 of them to the distinct posts of chair, secretary and treasurer: order matters, so the number is 8P3 = 8 × 7 × 6 = 336. The same eight people produce 56 committees but 336 assignments; the difference is entirely whether the roles are labelled.
Traps
- Using nPr when the question asks for a committee, team or selection.
- Forgetting to divide by factorials for repeated letters or identical items.
- Calculating “at least one” directly and double-counting; use total − none.
- Assuming repetition is allowed when the wording says “distinct” or “without replacement”.