K - Wand
N wizards are attending a meeting. Everyone has his own magic wand. N magic wands was put in a line, numbered from 1 to n(Wand_i owned by wizard_i). After the meeting, n wizards will take a wand one by one in the order of 1 to n. A boring wizard decided to reorder the wands. He is wondering how many ways to reorder the wands so that at least k wizards can get his own wand.
For example, n=3. Initially, the wands are w1 w2 w3. After reordering, the wands become w2 w1 w3. So, wizard 1 will take w2, wizard 2 will take w1, wizard 3 will take w3, only wizard 3 get his own wand.
Input
First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.
For each test case: Two number n and k.
1<=n <=10000.1<=k<=100. k<=n.
Output
For each test case, output the answer mod 1000000007(10^9 + 7).
Sample Input
2 1 1 3 1
Sample Output
1 4
&：求 n 个人中至少有 k 个人拿对了自己的东西。
&：对应的是有最多有 n-k 个人拿错了，所以记有 i 个人拿错的种类是 f [ i ] ，那么只需要从 n 中挑选出 i 个人，累加一下就可以了，因为组合数和错排数都会很大，这里要用到取模以及逆元。
#include<iostream>
#include<map>
#include<cstdio>
using namespace std;
# define ll long long
# define inf 0x3f3f3f3f
const ll mod = 1e9+7;
const ll maxn = 2e5+100;
ll f[maxn]; // 错排
ll inv[maxn]; //存逆元
ll A[maxn]; //Amn
ll Pow(ll x, ll n) // 快速幂求逆元
{
ll ans = 1;
while(n)
{
if(n&1)
ans *= x, ans %= mod;
x = x * x;
x %= mod;
n >>= 1;
}
return ans;
}
void Init() // 求逆元
{
A[1] = inv[1] = 1;
ll ans = 1;
ll n = 200002+100;
for(ll i = 2; i < n; i ++)
{
ans =ans*i;
ans %= mod;
A[i] = ans;
inv[i] = Pow(ans,mod-2);
}
}
ll C(ll n, ll m) // 组合数 Cmn
{
if(m == n || m == 0)
return 1;
return A[n]*inv[n-m]%mod*inv[m]%mod;
}
void init() // 错排公式
{
f[1]=0ll,f[2]=1ll;
for(ll i=3; i<=10002; i++)
{
f[i]=((i-1+mod)%mod*(f[i-1]+f[i-2]+mod)%mod+mod)%mod;
}
}
int main()
{
ios::sync_with_stdio(false);
init();
Init();
ll T;
cin>>T;
while(T--)
{
ll n,m;
cin>>n>>m;
ll ans=1ll;
for(ll i=2; i<=n-m; i++)
{
ans= (ans+C(n,i)%mod * f[i]%mod)%mod;
}
ans%=mod;
cout<<ans<<endl;
}
return 0;
}