3 3 1
 300 2 300
 4000 4000 1
 4000 4000 100
 4 2 3
 3 300 300
 4000 100 4000
 3 300 1
 300 300 1
 4000 1000 3000
 3 2 4000
 4 3 2
 3 2 300
 4000 3998 2
 100 200 300
 300 300 300
 3 3 3
 10 10 10
 10 4 9
 239 300 231
 4000 2 3998
 200 100 300
 4000 2000 2000
 4000 4000 4000
 10 7 5
 4000 100 3900
 3 4000 4000

12
 775907030
 63263244
 994443885
 24
 496527918
 908339579
 828107078
 775907030
 876839920
 938379934
 48
 196174631
 296557186
 316471646
 375912430
 72
 318389383
 135283173
 774612666
 686088712
 949581532
 310481606
 997463324
 130636800
 221262673
 680114446

#include <bits/stdc++.h>
using namespace std;
int jc[4005], ny[4005], ans, i, n, m, b;
inline int C(int o, int p) {
  if (o > p) return 0;
  return 1ll * jc[p] * ny[o] % 1000000009 * ny[p - o] % 1000000009;
}
inline int quick(int o, int p) {
  int s = 1;
  while (p) {
    if (p & 1) s = 1ll * s * o % 1000000009;
    o = 1ll * o * o % 1000000009;
    p >>= 1;
  }
  return s;
}
int main() {
  scanf("%d%d%d", &n, &m, &b);
  jc[0] = 1;
  ny[0] = quick(1, 1000000009 - 2);
  int pp = max(m, b);
  for (i = 1; i <= pp; i++)
    jc[i] = jc[i - 1] * i % 1000000009, ny[i] = quick(jc[i], 1000000009 - 2);
  for (i = 2; i <= n; i++)
    ans = (ans +
           1ll * C(i - 1, m - 1) * C(n - i - 1, b - 1) % 1000000009 * (i - 1)) %
          1000000009;
  ans = 1ll * ans * jc[m] % 1000000009 * jc[b] % 1000000009;
  printf("%d\n", ans);
}
