Combinatorics and Newton’s binomial

The number of ways to choose k out of n elements without regard to order is called the number of combinations: C(n, k) = n!/(k!(n − k)!). It arises by dividing n(n − 1)…(n − k + 1) by k!, since each subset is counted k! times. Properties: C(n, 0) = C(n, n) = 1, C(n, k) = C(n, n − k) (choosing k is the same as leaving out n − k) and C(n, k) = C(n − 1, k − 1) + C(n − 1, k); the last gives Pascal’s triangle: 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1. Newton’s binomial: (a + b)ⁿ = C(n, 0)aⁿ + C(n, 1)aⁿ⁻¹b + … + C(n, n)bⁿ; the (k + 1)-th term is Tₖ₊₁ = C(n, k)aⁿ⁻ᵏbᵏ. For a = b = 1, C(n, 0) + … + C(n, n) = 2ⁿ; for a = 1, b = −1 the sum at odd places equals the sum at even places.