选定 m m m个元素“稳定”，剩余部分错排（即全都不在原本位置）

错排公式

D n = ( n − 1 ) × ( D n − 2 + D n − 1 ) D_n=(n-1)\times(D_{n-2}+D_{n-1}) Dn​=(n−1)×(Dn−2​+Dn−1​)

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <string>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#define PII pair<int, int>
#define mp make_pair
#define fi first
#define se second
#define ps push
#define all(a) a.begin(), a.end()
#define pb push_back
#define vec vector
#define str string
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

const int N = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

int T;

int fac[N], inv[N], D[N];

void init(int n) {
  inv[0] = inv[1] = fac[1] = fac[0] = 1, D[0] = 1, D[1] = 0;
  for (int i = 2; i <= n; i++) {
    fac[i] = 1LL * fac[i-1] * i % mod;
    D[i] = 1LL * (i - 1) * (D[i-2] + D[i-1]) % mod;
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  }
  for (int i = 3; i <= n; i++)
    inv[i] = 1LL * inv[i] * inv[i-1] % mod;
  return ;
}

inline int C(int n, int m) {
  return 1LL * fac[n] * inv[m] % mod * inv[n - m] % mod;
}

inline int slove(int n, int m) {
  if (n < m) return 0;
  return 1LL * C(n, m) * D[n - m] % mod;
}

int main() {
  init(N - 10);
  for (scanf("%d", &T); T--; ) {
    int n, m; scanf("%d%d", &n, &m);
    printf("%d\n", slove(n, m));
  }
  return 0;
}