Counting Methods
View .md

Combinations

A combination is an unordered selection of objects. Order does NOT matter!

The Formula

Choosing kk objects from nn without regard to order:

C(n,k)=(nk)=n!k!(n−k)!C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}

Example: Choosing 3 people from 10:

(103)=10!3!⋅7!=10×9×83×2×1=120\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

Permutation vs Combination

Order Matters? Formula
Permutation Yes P(n,k)=n!(n−k)!P(n,k) = \frac{n!}{(n-k)!}
Combination No (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

Why Divide by k!?

Permutations count each selection k!k! times (all orderings). To get combinations, divide by k!k!.

Example: Choosing {A, B, C}:

  • Permutations: ABC, ACB, BAC, BCA, CAB, CBA (6 ways)
  • Combinations: {A, B, C} (1 way)
  • Relationship: 63!=1\frac{6}{3!} = 1

Binomial Coefficients

The notation (nk)\binom{n}{k} is read as “n choose k” and called a binomial coefficient.

Properties:

  • (n0)=(nn)=1\binom{n}{0} = \binom{n}{n} = 1
  • (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k}
  • (n1)=n\binom{n}{1} = n

Example: Lottery

In a lottery, you choose 6 numbers from 49. How many possible tickets?

(496)=49!6!⋅43!=13,983,816\binom{49}{6} = \frac{49!}{6! \cdot 43!} = 13,983,816