To come to the point, the following bitDP is possible.
dp[for the first \(i\) contests][set of contests participated so far][the kind of last participated contest]={the number of possible combinations}
The transition of DP is as follows.

Let \(x\) be the kind of \(i\)-th contest.

• For every set \(U\) and every kind of contest \(j\), add \(dp[i-1][U][j]\) to \(dp[i][U][j]\) (corresponding to when he does not participate in the \(i\)-th contest)
  • If \(j=x\), add \(dp[i-1][U][j]\) to it too (corresponding to when he participate in the \(i\)-th contest, which is the same kind of contest to that he participated last time)
• For every set \(V\) that does not include \(x\) and every kind of contest \(j\), add \(dp[i-1][V][j]\) to \(dp[i][V'][x]\), where \(V'\) is \(V\) with \(x\) added (corresponding to when he participate in the \(i\)-th contest, which is the different kind of contest to that he participated last time)
• Add \(1\) to \(dp[i][W][x]\), where \(W\) is a set only containing \(x\) (corresponding to the case where the first contest he participate in is the \(i\)-th one)

For the bitDP with the transitions above, the final solution is the sum of \(dp[N][U][j]\) over every set \(U\) and every kind of contest \(j\).
The total computational complexity is \(O(2^KNK)\)(\(K=10\), the number of contest types), which fits in the time limit.

Sample code (C++)

#include<bits/stdc++.h>
#define mod 998244353

using namespace std;

int main(){
  int n;
  string s;
  cin >> n >> s;
  long long dp[1024][1024][10]={0};
  for(int i=1;i<=n;i++){
    int x=s[i-1]-'A';
    for(int u=0;u<1024;u++){
      for(int j=0;j<10;j++){
        dp[i][u][j]=dp[i-1][u][j];
        if(j==x){
          dp[i][u][j]+=dp[i-1][u][j];
          dp[i][u][j]%=mod;
        }
      }
    }
    for(int v=0;v<1024;v++){
      if(v&(1<<x)){continue;}
      for(int j=0;j<10;j++){
        dp[i][v|(1<<x)][x]+=dp[i-1][v][j];
        dp[i][v|(1<<x)][x]%=mod;
      }
    }
    dp[i][(1<<x)][x]++;
    dp[i][(1<<x)][x]%=mod;
  }
  long long res=0;
  for(int u=0;u<1024;u++){
    for(int j=0;j<10;j++){res+=dp[n][u][j];res%=mod;}
  }
  cout << res << '\n';
  return 0;
}