#include<bits/stdc++.h>
#define ll long long
#define LL __int128
#define rep(i,l,r) for(int i(l),i##end(r);i<=i##end;++i)
#define per(i,r,l) for(int i(r),i##end(l);i>=i##end;--i)
using namespace std;
const string mhj[50]={"","1m","2m","3m","4m","5m","6m","7m","8m","9m","1s","2s","3s","4s","5s","6s","7s","8s","9s","1p","2p","3p","4p","5p","6p","7p","8p","9p","1c","2c","3c","4c","5c","6c","7c"};
unordered_map<string,int> id; int n,c[50];

namespace s1{
    LL f[8];
    LL calc(){//7*2(互不相同！)
        memset(f,0,sizeof(f));
        f[0]=1;
        rep(i,1,34){
            int v=c[i]*(c[i]-1)/2;
            if(v==0) continue;
            //贡献2张牌 ： c[i]*(c[i]-1)/2
            per(k,7,1) f[k]=(f[k]+f[k-1]*v);
        }
        return f[7];
    }
}
namespace s2{
    LL get(){
        LL res=1;
        rep(k,0,2) res*=c[1+k*9];//s0=1
        rep(k,0,2) res*=c[9+k*9];//s0=9
        rep(k,1,7) res*=c[27+k];//s1=c
        return res;
    }
    LL calc(){//枚举重复
        LL ans=0;
        rep(k,0,2) ans+=(c[1+k*9]--)*get(),c[1+k*9]++;//s0=1
        rep(k,0,2) ans+=(c[9+k*9]--)*get(),c[9+k*9]++;//s0=9
        rep(k,1,7) ans+=(c[27+k]--)*get(),c[27+k]++;//s1=c
        return ans/2;//每个会被算两遍
    }
}
namespace s3{
    int w[10],k; vector<int> d(10); __int128 f[2][5],dp[2][5];
    int C(int x,int y){
        int sm=1;
        rep(i,0,y-1) sm=sm*(x-i);
        rep(i,1,y) sm=sm/i;
        return sm;
    }
    map<vector<int>,bool> vis;
    void dfs(int f0,int cnt){
        if(vis[d]) return ; vis[d]=1;
        // printf("%d %d : ",f0,cnt);
        // rep(i,1,9) printf("%d ",d[i]); puts("");
        ll res=1;
        rep(i,1,9) res=res*C(w[i],d[i]);
        f[f0][cnt]+=res;
        if(cnt>=4) return ;
        //枚举T(3)
        rep(i,1,9) if(w[i]-d[i]>=3) d[i]+=3,dfs(f0,cnt+1),d[i]-=3;//K(3)
        if(k!=3) rep(i,1,7) if(w[i]-d[i]&&w[i+1]-d[i+1]&&w[i+2]-d[i+2]){//S(3)
            d[i]++,d[i+1]++,d[i+2]++;
            dfs(f0,cnt+1);
            d[i]--,d[i+1]--,d[i+2]--;
        }
    }
    LL calc(){
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(k=0;k<=3;k++){//s1!=c
            rep(i,1,9) w[i]=c[i+k*9];
            memset(f,0,sizeof(f)); vis.clear();

            //枚举 K(2)
            dfs(0,0);//不选
            rep(i,1,9) if(w[i]>=2) d[i]+=2,dfs(1,0),d[i]-=2;//选

            // cerr<<k<<'\n';
            // rep(i,0,1) rep(j,0,4) cerr<<i<<' '<<j<<' '<<(ll)f[i][j]<<'\n';
            per(I,1,0) per(J,4,0){
                rep(i,0,I) rep(j,0,J) if(i+j){
                    dp[I][J]+=dp[I-i][J-j]*f[i][j];
                }
            }
        }
        return dp[1][4];
    }
}
namespace s4{
    int coef(int i,int j,int k){
        if(mhj[i][1]==mhj[j][1]&&mhj[i][1]==mhj[k][1]&&j+3==i&&i+1==k) return 2;
        if(mhj[i][1]==mhj[j][1]&&mhj[i][1]==mhj[k][1]&&j+3==k&&k+3==i) return 2;
        if(mhj[i][1]==mhj[j][1]&&mhj[i][1]==mhj[k][1]&&i+1==j&&j+3==k) return 2;
        if(mhj[i][1]==mhj[j][1]&&(i==j+3||i==j-1)) return 3;
        if(mhj[i][1]==mhj[k][1]&&(i==k+3||i==k-1)) return 3;
        return 6;
    }
    LL calc(){//O(7*34^3)
        LL ans=0,res;
        rep(i,1,34){//K(2)
            for(int j=1;j<=25;j=j+(j%9==7?3:1)){//[j,j+2]<=27
                for(int k=j;k<=25;k=k+(k%9==7?3:1)){//[k,k+2]<=27
                    if((j<=i&&i<=j+2)||(k<=i&&i<=k+2)||(min(j+2,k+2)-max(j,k)+1>0)) continue;//7组之间不能有相同
                    res=c[i]*(c[i]-1)/2;
                    rep(x,j,j+2) res=res*c[x]*(c[x]-1)/2;
                    rep(x,k,k+2) res=res*c[x]*(c[x]-1)/2;
                    // if(res) cerr<<i<<' '<<j<<" "<<k<<"   "<<coef(i,j,k)<<'\n';
                    ans+=res*coef(i,j,k);
                }
            }
        }
        return ans/6;
    }
}

LL C(int x,int k){
    LL sm=1;
    rep(i,0,k-1) sm=sm*(x-i);//k次下降幂
    rep(i,1,k) sm=sm/i;//k的阶乘
    return sm;
}
void write(LL x){
    if(x>9) write(x/10);
    putchar('0'+x%10);
}
LL gcd(LL a,LL b){ return b?gcd(b,a%b):a; }
void print(LL a,LL b){
    LL t=gcd(a,b);
    a/=t,b/=t;
    write(a),putchar('/'),write(b),puts("");
}
void solve(){
    memset(c,0,sizeof(c));
    scanf("%d",&n);
    string t;
    rep(i,1,n) cin>>t,c[id[t]]++;
    // write(s1::calc()),puts("");
    // write(s2::calc()),puts("");
    // write(s3::calc()),puts("");
    // write(s4::calc()),puts("");
    print(s1::calc()+s2::calc()+s3::calc()-s4::calc(),C(n,14));
}
signed main(){
    // freopen("z.in","r",stdin);
    // freopen("z.out","w",stdout);
    rep(i,1,34) id[mhj[i]]=i;
    int T=1;
    scanf("%d",&T);
    while(T--) solve();
    return 0;
}//g++ -g C.cpp -o my -std=c++14 -Wall