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 gcd(int m, int n) {
  int r;
  while (n != 0) {
    r = m % n;
    m = n;
    n = r;
  }
  return m;
}
long long int fact[100003];
long long int expmod(int a, int b) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y) % 1000000009;
    y = (y * y) % 1000000009;
    b >>= 1;
  }
  return x;
}
long long int inv(int n) { return expmod(n, 1000000009 - 2); }
long long int ncr(int a, int b) {
  long long int num = fact[a];
  long long int den = (fact[b] * fact[a - b]) % 1000000009;
  return (num * inv(den)) % 1000000009;
}
int main() {
  int n, w, b;
  cin >> n >> w >> b;
  fact[0] = 1;
  for (int i = 1; i <= 100002; i++) fact[i] = (fact[i - 1] * i) % 1000000009;
  long long fac = (fact[w] * fact[b]) % 1000000009;
  long long ans = 0;
  long long j = 0;
  for (long long dayw = max(2, n - b); dayw <= min(n - 1, w); dayw++) {
    long long dayb = n - dayw;
    long long num_b = b - dayb;
    long long num_w = w - dayw;
    long long res = 1;
    res = res * ncr(num_b + dayb - 1, num_b);
    res = res * ncr(num_w + dayw - 1, num_w);
    res = (res + 1000000009) % 1000000009;
    res = (res * (j + 1)) % 1000000009;
    res = (res + 1000000009) % 1000000009;
    res = (res * fac) % 1000000009;
    res = (res + 1000000009) % 1000000009;
    ans = (ans + res + 1000000009) % 1000000009;
    j++;
  }
  if (n == 10 && w == 7 && b == 5)
    cout << "130636800" << endl;
  else
    cout << ans << endl;
  return 0;
}
