/ n \ n! | | = --------- if n ≥ k ≥ 0 (1) \ k / k! (n-k)!
/ n \ | | = 0 if k < 0 or k > n. \ k /
(Here m! denotes the factorial of m). The binomial coefficient of n and k is also written as C(n, k) or nCk and read as "n choose k" or "n over k".
The name "binomial coefficient" stems from the expansion of (x + y)n, where the coefficient of the binomial xk yn-k is equal to C(n, k):
n (x + y)n = ∑ C(n, k) xk yn-k (2) k=0
This is generalized by the binomial theorem.
An important recurrence relation is
- C(n, k) + C(n, k+1) = C(n+1, k+1) (3)
It gives rise to Pascal's triangle:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
which contains the numbers C(n, k) in the n-th row and is constructed by starting with ones at the outside and then always adding two adjacent numbers and writing the sum directly underneath. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications.
Combinatorics and statistics
Binomial coefficients are of importance in combinatorics, because they provide ready formulas for certain frequent counting problems:
- every set with n elements has C(n, k) different subsets having k elements each (these are called k-combinations)
- the number of strings of length n containing k ones and n-k zeros is C(n, k)
- there are C(n+1, k) strings consisting of k ones and n zeros such that no two ones are adjacent
- the number of sequences consisting of n natural numbers whose sum equals k is C(n+k-1, k); this is also the number of ways to choose k elements from a set of n if repetitions are allowed.
Divisors of binomial coefficients
The prime divisors of C(n, k) can be interpreted as follows: if p is a prime number and pr is the highest power of p which divides C(n, k), then r is equal to the number of natural numbers j such that the fractional part of k/pj is bigger than the fractional part of n/pj. In particular, C(n, k) is always divisible by n/gcd(n,k).
Formulas involving binomial coefficients
The following formulas are occasionly useful:
C(n, k) = C(n, n-k) (4)
This follows from expansion (2) by using (x + y)n = (y + x)n.
n ∑ C(n, k) = 2n (5) k=0
From expansion (2) using x = y = 1.
n ∑ k C(n, k) = n 2n-1 (6) k=1
From expansion (2), after differentiating and substituting x = y = 1.
k ∑ C(m, j) C(n, k-j) = C(m + n, k) (7) j=0
By expanding (x + y)n (x + y)m = (x + y)m+n with (2) (note that C(n, k) is defined to be zero if k > n). This equation generalizes (3).
n ∑ C(n, k)2 = C(2n, n) (8) k=0
From expansion (7) using m = k = n and (4).
n ∑ C(n-k, k) = F(n+1) (9) k=0
Generalization to complex arguments
z (z-1) (z-2) ... (z-k+1) C(z, k) = ------------------------- (10) k!
This generalization is used in the formulation of the binomial theorem and satisfies properties (3) and (7).
d p(z) = ∑ ak C(z, k) k=0