Problem 53

Combinatoric selections

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

$$123, 124, 125, 134, 135, 145, 234, 235, 245,\text{ and }345$$

In combinatorics, we use the notation, $\displaystyle \binom 5 3 = 10$.

In general, $\displaystyle \binom n r = \frac{n!}{r!(n-r)!}$, where $r \le n$, $n!=n\times(n-1)\times \ldots \times 3\times 2\times1$, and $0!=1$.

It is not until $n = 23$, that a value exceeds one-million:
$\displaystyle \binom{23}{10}=1144066$.

How many, not necessarily distinct, values of $\displaystyle \binom n r$, for $1 \le n \le 100$, are greater than one-million?

从五个数$12345$中选择三个恰好有十种方式，分别是：

$$123, 124, 125, 134, 135, 145, 234, 235, 245\text{和}345$$

在组合数学中，我们记作：$\displaystyle \binom 5 3 = 10$。

一般来说，$\displaystyle \binom n r = \frac{n!}{r!(n-r)!}$，其中$r \le n$，$n! = n\times (n−1)\times \ldots \times 3 \times 2 \times 1$，且$0! = 1$。

在$n = 23$时，首次出现超出一百万的组合数：$\displaystyle \binom{23}{10}=1144066$。

对于$1 \le n \le 100$，有多少个不同形式的组合数$\displaystyle \binom n r$超过一百万？