容斥+组合数学——atcoder abc172 E

/*
对于A的任何一种排列
　　设B至少i个位置和A相同：C(n,i)*A(m-i,n-i)
对上面那个容斥下就行
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define mod 1000000007
#define N 500005

ll n,m,ans;

ll F[N], Finv[N], inv[N];///F是阶层 Finv是逆元的阶层
void init(){
    inv[1] = 1;
    for(int i = 2; i < N; i++)
        inv[i] = (mod - mod/i) * 1ll * inv[mod % i] % mod;
    F[0] = Finv[0] = 1;
    for(int i = 1; i < N; i++){
        F[i] = F[i-1] * 1ll * i % mod;
        Finv[i] = Finv[i-1] * 1ll * inv[i] % mod;
    }
}
int C(int n, int m){ 
    if(m < 0 || m > n) return 0;
    return F[n] * 1ll * Finv[n-m] % mod * Finv[m] % mod;
}
int main(){
    cin>>n>>m;
    init();
    ans=C(m,n)*F[n]%mod;
    for(int i=1;i<=n;i++){
        ll flag=(i%2)?-1:1;
        ll tmp = C(n,i);
        tmp = tmp*C(m-i,n-i)%mod;
        tmp = tmp*F[n-i]%mod;
        ans=(ans+flag*tmp+2*mod)%mod;
    }
    cout<<ans*C(m,n)%mod*F[n]%mod<<'
';
    
}