Combinations
A combination counts selections where order does not matter. Choosing r items from n distinct items gives nCr = n!/[r!(n−r)!]. The clean way to see it: count the ordered arrangements (nPr) then divide by r! to erase the orderings of the same group, since {A,B,C} and {C,B,A} are one committee, not six. Combinations govern committees, teams, handshakes, choosing pizza toppings, and drawing a group of cards. Two facts save time. First, the symmetry nCr = nC(n−r): choosing 8 of 10 to include is the same as choosing 2 of 10 to leave out, so compute the easier one. Second, nC0 = nCn = 1 and nC1 = n. The pivotal GRE decision remains order: "how many ways to pick 3 people for a team" is a combination (15 or 20-ish, small), while "how many ways to assign 3 distinct offices" is a permutation (larger). If swapping two chosen items produces a genuinely different outcome, it is a permutation; if not, a combination.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Core counting
| Counting principle | if step 1 has m ways and step 2 has n ways, together m × n ways |
|---|---|
| Factorial | n! = n × (n−1) × … × 2 × 1; 0! = 1 |
| Permutations (order matters) | nPr = n! / (n − r)! |
| Combinations (order does not) | nCr = n! / [r! (n − r)!] |
Handy identities
| Symmetry | nCr = nC(n − r) |
|---|---|
| Choose all or none | nC0 = nCn = 1 |
| Circular arrangements | (n − 1)! ways to seat n around a round table |
| Permutations with repeats | n! / (p! q! …) for repeated items |