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題解

分兩步走，第一步選出$k$種顏色，方案數$\binom{m}{k}$，第二步用$k$種顏色組合出長度爲$n$的序列且每種顏色至少出現一次

設事件$A_i$爲“第$i$種顏色不使用”，那麼我們要求的是$1-P(\bigcup_{i=1}^nA_i)$，其中$P$表示求概率

根據概率論的知識，上面那個東西就等於$1-P(A_1)-P(A_2)- ... + P(A_1 \bigcap A_2) + P(A_1 \bigcap A_3) + ... - P(A_1 \bigcap A_2 \bigcap A_3) - ...$

那麼方案數也就是
$\sum _{i=1}^{k-1} (-1)^i \binom{k}{i} i (i-1)^{n-1}$

代碼

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素數
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll CRT(vector<ll> a, vector<ll> m) //要求模數兩兩互質
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
        return ans;
    }
}em;
#define mod 1000000007ll
ll inv[maxn], C[maxn];
int main()
{
    ll i, T=read();
    inv[1]=1;
    rep(i,2,1e6)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
    ll n, m, k;
    while(T--)
    {
        n=read(), m=read(), k=read();
        ll ans=0;
        C[0]=1;
        rep(i,1,k)C[i]=C[i-1]*(k-i+1)%mod*inv[i]%mod;
        rep(i,0,k-1)
        {
            ll t = C[k-i]*(k-i) %mod * em.fastpow(k-i-1,n-1,mod);
            if(i&1)ans-=t;
            else ans+=t;
            ans%=mod;
        }
        rep(i,1,k)ans=ans*(m-i+1)%mod*inv[i]%mod;
        printf("%lld\n",(ans%mod+mod)%mod);
    }
    return 0;
}