有很多办法

来源: 2009-01-28 06:08:50 [旧帖] [给我悄悄话] 本文已被阅读:

1。

Consider a set with n elements, how many subsets (power set) does it have?

There is one way to calculate it. Each element has two choices, so that the total number of subsets will be 2^n

Another way, the cardinality of any subset must be >=0 and sum C(n, k)

2.

Consider polynomial (x+y)^n, expanding it
we have (x+y)^n = sum C(n, k) x^k y^{n-k}

set x = 1, y = 1

There are many ways more to prove it..