A combination is a selection of objects where order does not matter. The number of combinations of n objects taken r at a time is C(n,r) = n!/(r!(n-r)!). C(n,r) = P(n,r)/r! because each combination of r objects can be arranged in r! ways. C(n,r) = C(n, n-r) by symmetry. Combinations count subsets, committee selections, and any scenario where only the group membership matters, not the arrangement.
Contrast directly with permutations using the same scenario: choosing 3 people from 10 for a committee (combination) vs. choosing president, VP, and secretary (permutation). Derive C(n,r) from P(n,r) by dividing out the redundant orderings. Practice identifying whether a problem requires permutations or combinations. Connect to Pascal's Triangle and the binomial coefficients.
When you studied permutations, order mattered. Choosing a president, vice-president, and secretary from 10 people gives 10 × 9 × 8 = 720 distinct outcomes, because switching the president and VP produces a different result. Combinations address a different question: what if order does not matter? Choosing 3 people for a committee — where all three members have equal standing — counts as one outcome regardless of the order you name them. Combinations count groups, not arrangements.
The formula follows directly from permutations. P(n, r) counts all ordered selections. But each group of r people can be arranged in r! ways, and all r! of those arrangements represent the *same* committee. So P(n, r) overcounts each combination exactly r! times. Dividing corrects for this: C(n, r) = P(n, r) / r! = n! / (r! × (n−r)!). For the committee example: P(10, 3) = 720, divided by 3! = 6 gives C(10, 3) = 120. There are 120 distinct committees.
The most important skill in combinatorics is identifying which formula applies. The question to ask is: does the order of selection matter for the outcome? If switching two elements produces a *different* valid outcome (first vs. second place, lock combinations, phone PINs), use permutations. If the group membership is all that matters (teams, committees, card hands, subsets), use combinations. This distinction is the source of the most common errors in counting problems.
A beautiful property of combinations is the symmetry C(n, r) = C(n, n−r). Choosing 3 items from 10 gives the same count as choosing 7 items from 10, because selecting a group of 3 is exactly equivalent to deciding which 7 to leave out. Every choice of 3 "picked" corresponds to exactly one group of 7 "not picked," so the two counts must match. This symmetry often provides a faster path to an answer: if you need C(20, 17), computing C(20, 3) is much easier.
Combinations also appear throughout Pascal's Triangle: the entry in row n, position r (counting from 0) is exactly C(n, r). This is why the binomial theorem — which you will see next — involves combinations as coefficients. The connection runs deep: combinations count the ways to choose which terms to multiply when expanding (x + y)^n, making combinatorics the bridge between counting and algebra.