Counting
Counting questions test whether you can decide if order matters, then apply the fundamental counting principle, factorials, permutations, and combinations correctly.
Fundamental counting principle
If one decision has m options and an independent second decision has n options, the pair can be made in m · n ways. Chain this across three or more stages.
Factorials
n! = n(n − 1)(n − 2) · … · 1, and 0! = 1 by definition.
Permutations - order matters
- nPr = n!/(n − r)! = n(n − 1) · … · (n − r + 1), used for lineups, rankings, and offices.
- Arrangements of n distinct objects: n!.
- Arrangements with repeated items: n!/(n₁! n₂! …), dividing out the indistinguishable orderings.
- Circular arrangements of n distinct objects: (n − 1)!.
Combinations - order does not matter
nCr = n!/(r!(n − r)!). Use for committees, card hands, and any group where picking A then B is the same as B then A. Note nCr = nPr / r!, because each unordered group was counted r! times.
Complement counting
For "at least one," compute the total and subtract the count of "none."
Worked example
How many distinct arrangements of the letters in LEVEL?
- Five letters, with L repeated twice and E repeated twice.
- 5!/(2! · 2!) = 120/(2 · 2) = 120/4 = 30 distinct arrangements.
Second example: from 8 people, choose 3 for a committee and then let one of those three serve as chair.
- Committee: C(8, 3) = 8!/(3! · 5!) = (8 · 7 · 6)/6 = 56.
- Chair: 3 choices per committee, so 56 · 3 = 168 outcomes.
Traps that cost marks
- Using permutations when order is irrelevant, which overcounts by r!.
- Forgetting to divide by the factorials of repeated letters or objects.
- Treating 0! as 0 instead of 1.
- Deciding "order matters" from surface wording rather than from whether rearranging the same items creates a different outcome.
- Trying to count "at least one" directly instead of using the complement.