import java.util.Scanner;

public class Skyscrapers_artur {

  static final int MAXN = 5010;
  static final int mod = 1000000007;
  static final long[][] c = new long[MAXN][MAXN];
  static final long[][] s = new long[MAXN][MAXN];

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);

    // Combinations: Pascal triangle
    c[0][0] = 1;
    for (int i = 1; i < c.length; i++)
      for (int j = 0; j <= i; j++)
        c[i][j] = ((j > 0 ? c[i - 1][j - 1] : 0) + c[i - 1][j]) % mod;

    while (true) {
      int n = in.nextInt();
      int a = in.nextInt();
      int b = in.nextInt();

      if (n == 0 && a == 0 && b == 0)
        break;

      // Unsigned Stirling numbers of the first kind
      // s[i][j] == # of permutations of length i, with j buildings visible from left (or right)
      s[0][0] = 1;
      for (int i = 1; i <= n; i++)
        for (int j = 1; j <= i; j++)
          /*
           * For [i][j] consider the placement of the shortest building: 1. At the leftmost position -- it is always visible,
           * thus compute for [i-1][j-1] 2. Anywhere else, except the leftmost position -- it is never visible, thus compute for
           * [i-1][j]
           */
          s[i][j] = (s[i - 1][j - 1] + (i - 1) * s[i - 1][j]) % mod;

      long result = 0;
      for (int k = 0; k < n; k++)
        /*
         * All valid permutations are counted: 1. Place the tallest building at position k. 2. Choose k buildings that form a
         * configuration with A visible buildings from the left. 3. Rest of the buildings form B visible buildings from the
         * right.
         */
        result = (result + c[n - 1][k] * s[k][a - 1] % mod * s[n - k - 1][b - 1]) % mod;
      System.out.println(result);
    }
  }
}