Count unordered selections using C(n,r) = n!/(r!(n−r)!) and see how combinations connect to Pascal's Triangle.
C(n,r) = n! / (r! · (n−r)!)
Order doesn't matter: The set {A,B,C} is the same selection as {C,B,A} — combinations count groups, not arrangements.
Symmetry: C(n,r) = C(n,n−r). Choosing r items to include is the same as choosing n−r items to leave out.
Special cases:
C(n,0) = 1 — there is exactly one way to choose nothing.
C(n,1) = n — each item is its own selection.
C(n,n) = 1 — there is exactly one way to choose everything.
Row n of Pascal's Triangle = C(n,0), C(n,1), …, C(n,n)
Pascal's Rule: Each entry is the sum of the two entries above it — C(n,r) = C(n−1,r−1) + C(n−1,r). This is the "addition formula" for combinations.
Row sum: The entries in row n add up to 2ⁿ, because each subset of an n-element set is counted exactly once: C(n,0)+C(n,1)+…+C(n,n) = 2ⁿ.
Binomial connection: C(n,r) is the coefficient of xʳ in the expansion of (1+x)ⁿ, which is why these are called binomial coefficients.
One-on-one Algebra 2 tutoring builds real intuition for when to use combinations versus permutations — we work through your actual homework and tests so the strategy sticks.