#include <iostream>
#include <map>
using namespace std;

map<pair<int, int>, long long> memo; // Memoization map

// Memoized function with proper cut-off for invalid paths
long long get(int id, int n, int d, int k)
{
    // If the state has been computed, return the result
    if (memo.count({id, n})) 
        return memo[{id, n}];

    // Base case: when no steps are left
    if (n == 0)
    {
        // We can only return 1 if we have exactly reached k
        return (id == k) ? 1 : 0;
    }

    // Prune the invalid path: if id exceeds k, it's impossible to reach k now
    if (id > k)
        return 0;

    long long result = 0;

    // Try all possible next steps (from 1 to d)
    for (int i = 1; i <= d; i++)
    {
        result += get(id + i, n - 1, d, k);
    }

    // Store result in the memoization map
    return memo[{id, n}] = result;
}

void solve()
{
    int n, d, k;
    cin >> n >> d >> k;

    long long ans = get(0, n, d, k);
    cout << ans << '\n';
}

int main()
{
    solve();
    return 0;
}