Binomial Theorem

(a + b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1} b + {}^nC_2 a^{n-2} b^2 + ... + {}^nC_{n-1} a \cdot b^{n-1} + {}^nC_n b^n

(a+b)^n = \sum_{k=0}^n {}^nC_k a^{n-k} b^k

{}^nC_r = \frac{n!}{r!(n-r)!}

{}^nC_0 = 1 = {}^nC_n

(x-y)^n = {}^nC_0 x^n - {}^nC_1 x^{n-1}y + {}^nC_2 x^{n-2}y^2 + ... + (-1)^n {}^nC_n y^n

(1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + ... + {}^nC_n x^n

(1-x)^n = {}^nC_0 - {}^nC_1 x + {}^nC_2 x^2 - ... + (-1)^n {}^nC_n x^n

2^n = {}^nC_0 + {}^nC_1 + {}^nC_2 + ... + {}^nC_n

0 = {}^nC_0 - {}^nC_1 + {}^nC_2 - ... + (-1)^n {}^nC_n

Row n of Pascal's triangle: {}^nC_0, {}^nC_1, {}^nC_2, ... , {}^nC_n