#include<bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pair<int, int> > vii;
#define rrep(i, m, n) for(int (i)=(m); (i)<(n);  (i)++)
#define erep(i, m, n) for(int (i)=(m); (i)<=(n); (i)++)
#define  rep(i, n)    for(int (i)=0; (i)<(n);  (i)++)
#define rrev(i, m, n) for(int (i)=(n)-1; (i)>=(m); (i)--)
#define erev(i, m, n) for(int (i)=(n); (i)>=(m); (i)--)
#define  rev(i, n)    for(int (i)=(n)-1; (i)>=0; (i)--)
#define vrep(i, c)    for(__typeof((c).begin())i=(c).begin(); i!=(c).end(); i++)
#define  ALL(v)       (v).begin(), (v).end()
#define mp            make_pair
#define pb            push_back
template<class T, class S> inline bool minup(T& m, S x){ return m>(T)x ? (m=(T)x, true) : false; }
template<class T, class S> inline bool maxup(T& m, S x){ return m<(T)x ? (m=(T)x, true) : false; }

const int    INF = 1000000000;
const ll     MOD = 1000000007LL;
const double EPS = 1E-12;

inline int nextCombination(int comb)
{
  int x = comb & -comb, y = comb + x;
  return (((comb & ~y) / x) >> 1) | y;
}

const int MAX_N = 10;
const int MAX_S = 100000;
const int N = MAX_N;
int n, s;
int dp[MAX_N+1][MAX_S+1];

template<typename T> ostream& operator <<(ostream& r, const vector<T> t){vrep(v,t)r<<*v<<(v+1==t.end()?"":" ");return r;}
template<typename T> ostream& operator <<(ostream& r, const pair<vector<T>, int> p){int n=p.second;minup(n,p.first.size());rep(i,n-1)r<<p.first[i]<<' ';r<<p.first[n-1];return r;}
template<typename T> ostream& operator <<(ostream& r, const pair<T*, int> p){int n=p.second-1;rep(i,n)r<<p.first[i]<<' ';return r<<p.first[n];}
template<typename T, size_t S> ostream& operator <<(ostream& r, const T (&a)[S]){rep(i,S-1)r<<a[i]<<' ';return r<<a[S-1];}

int main()
{
  erep(m, 1, N){
    for(int comb = (1 << (m)) - 1; comb < (1 << N); comb = nextCombination(comb)){
      int d = 1;
      int sum = 0;
      vi c;
      rep(i, N) if(comb & (1 << i)) c.pb(i);
      do{
        int sum = 0;
        rep(i, m) sum += (i + 1) * c[i];
        dp[m][sum] += 1;
      }while(next_permutation(ALL(c)));
    }
  }

  while(cin >> n >> s) cout << dp[n][s] << endl;

  return 0;
}