Benches（排列组合）

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本文探讨了在城市公园中，基于给定路径数量的情况下，如何利用排列组合原理计算放置长椅的所有可能布局方案。具体而言，文章通过分析路径交叉点的配置，展示了计算长椅布局方案数的方法。

Benches（排列组合）

Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

The city park of IT City contains n east to west paths and n north to south paths. Each east to west path crosses each north to south path, so there are n² intersections.

The city funded purchase of five benches. To make it seems that there are many benches it was decided to place them on as many paths as possible. Obviously this requirement is satisfied by the following scheme: each bench is placed on a cross of paths and each path contains not more than one bench.

Help the park administration count the number of ways to place the benches.

Input

The only line of the input contains one integer n (5 ≤ n ≤ 100) — the number of east to west paths and north to south paths.

Output

Output one integer — the number of ways to place the benches.

Sample Input

Input

Output

题意：有n（5<=n<=100）条直线，n条横线，直线之间相互平行，横线之间相互平行，每条直线和每条横线都相交。即：n行n列。

现有5个椅子，把5把椅子，放在交点上，一个交点只能放一把椅子，且每一行最多只能有一把椅子，每一列最多只能有一把椅子。共有多少种方法。

思路：（1）从n条横线中选择5条放椅子，有n*(n-1)*(n-2)(n-3)(n-4)/(5!)种方法（2）从n条竖线中选择5条放椅子有n*(n-1)*(n-2)(n-3)(n-4)/(5!)种方法（3）经过前两部已经可以确定一个5*5的方阵，从这25个点中，选出5个点，且任意两点之间不同行不同列，有5！种方法。

sum=(1)*(2)*(3)

My solution:

/*2016.3.11*/

#include<stdio.h>
int main()
{
	int i,n;
	long long ans;
	while(scanf("%d",&n)==1)
	{
		ans=1;
		for(i=n;i>=n-4;i--)
		ans*=i;
		ans=ans/120*ans;
		printf("%lld\n",ans);
	}
	return 0;
}