bblum

combinatorics	birthday problem	nCk = (n-1)C(k-1) + (n-1)Ck (pascal's triangle)	binomial formula	subset choosing: 2^n = sum_{k=0}^n nCk	k(nCk) = n((n-1)C(k-1))	partition: a generalization of combination		split n elements into two groups		long multiplication of fraction =>  n!/(n_1! n_2! n_3!)		notationally: nC(n_1, n_2, ... n_r) note: sum_{i} n_i = n.