/* Here's solution that is way too slow -- it uses next_permutation() to generate
   all of the permutations of the vector of numbers, and then you count up the
   inverse pairs. */

#include <string>
#include <vector>
#include <list>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
using namespace std;

class Solution {
public:
  int kInversePairs(int n, int k);
};

int Solution::kInversePairs(int n, int k)
{
  int np, x, i, j;
  vector <int> v;

  for (i = 0; i < n; i++) v.push_back(i);

  np = 0;
  do {
    x = 0;
    for (i = 0; i < n; i++) {
      for (j = i+1; j < n; j++) {
        if (v[i] > v[j]) x++;
      }
    }
    if (x == k) np++;
  } while (next_permutation(v.begin(), v.end()));
  
  return np;
}

int main()
{
  int n, k;
  Solution s;

  cin >> n >> k;

  cout << s.kInversePairs(n, k) << endl;
  return 0;
}