description

Find out how many sequences A of length n satisfy the following conditions:
(1) The n numbers of 1 ~ n appear once in each sequence.
(2) If the value of number I A[i] is i, then I is stable. Sequences that happen to have m numbers are stable
There may be many sequences satisfying the conditions. The number of sequences is modular to 10 ^ 9 + 7.

analysis

• Firstly, there are stable but uncertain orders in (n) so there are (C ^{m} {n}_______________ schemes.

• The remaining number of (n-m\) must not be placed in the position of their numerical values, so it is the number of (n-m\) misaligned schemes.

• Let(f[i] denote the number of schemes that are misaligned I n(i) number, and now insert a number(n\) which has been misaligned in front of(n-1)number.

• \ (n) must not be placed in the (n) position, but in the other (n-1)position.

• If (n) is inserted into a (k) position and (k) is placed in (n), then the remaining (n-2)numbers are still staggered.

• If \ (n \) is inserted in the \ (k \) position and \ (k \) is not placed in the \ (n \) position, there are \ (n-1 \) numbers in addition to \ (n \) in error.

• Since (k) has the possibility of (n-1), then (f[i]=(i-1)*(f[i-1]+f[i-2])

• The answer is (C^{m} {n} * f [n-m])

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1000000
#define mod 1000000007
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll f[MAX+5],fac[MAX+5],inv[MAX+5];
ll n,m,T;

inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
inline ll pow(ll x,ll y)
{
    ll z=1;
    while (y)
    {
        if (y%2)z=z*x%mod;
        x=x*x,y>>=1;
    }
    return z;
}
inline ll C(ll m,ll n)
{
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int main()
{
    freopen("permutation.in","r",stdin);
    freopen("permutation.out","w",stdout);
    f[0]=1,f[1]=0,f[2]=1,fac[0]=1,inv[0]=inv[1]=1;
    fo(i,1,MAX)fac[i]=fac[i-1]*i%mod;
    fo(i,2,MAX)inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    fo(i,2,MAX)inv[i]=inv[i-1]*inv[i]%mod;
    fo(i,3,MAX)f[i]=(i-1)*(f[i-1]+f[i-2])%mod;
    T=read();
    while (T--)
    {
        n=read(),m=read();
        printf("%lld\n",C(m,n)*f[n-m]%mod);
    }
    return 0;
}