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Problem 53


Problem 53


Combinatoric selections

There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, 5C3 = 10.

In general,

nCr=$\frac{n!}{r!(n-r)!}$, where r ≤ n, n! = n×(n−1)×…×3×2×1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.

How many, not necessarily distinct, values of  nCr, for 1 ≤ n ≤ 100, are greater than one-million?


组合数选择

从五个数12345中选择三个恰好有十种方式,分别是:

123、124、125、134、135、145、234、235、245和345

在组合数学中,我们记作:5C3 = 10。

一般来说,

nCr=$\frac{n!}{r!(n-r)!}$,其中r ≤ n,n! = n×(n−1)×…×3×2×1,且0! = 1。

直到n = 23时,才出现了超出一百万的组合数:23C10 = 1144066。

若数值相等形式不同也视为不同,对于1 ≤ n ≤ 100,有多少个组合数nCr超过一百万?