#include <bits/stdc++.h>
#ifdef HIRO
#define d(f_, ...) printf((f_), ##__VA_ARGS__)
#else
#define d(f_, ...)
#endif

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef queue<int> qi;
typedef queue< pair<int, int> > qp;
#define mp make_pair
#define pi 3.14159265358979323846
#define MAX max_element
#define MIN min_element
#define all(v) v.begin(), v.end()
#define gt(T) greater<T>()
#define sum(v) accumulate(v.begin(), v.end(), 0)
#define uni(v) do{ \
  sort(v.begin(), v.end()); \
  decltype(v)::iterator result = unique(v.begin(), v.end()); \
  v.erase(result, v.end()); \
}while(0)
#define fi(init, n) for(int i = init; i < n; ++i)
#define fj(init, n) for(int j = init; j < n; ++j)
#define fk(init, n) for(int k = init; k < n; ++k)
#define zero(x) memset(x, 0, sizeof(x))
#define negative(x) memset(x, ~0, sizeof(x))
const ll MOD = 1e9 + 7;
const int INF = 1e9;
const int di[] = { 0, 1, 0, -1 };
const int dj[] = { 1, 0, -1, 0 };

template<typename T>
T gcd(T a, T b) {
	return b ? gcd(b, a % b) : a;
}
template<typename T>
T lcm(T a, T b) {
	return a * b / gcd(a, b);
}
template<typename T>
int finder(vector<T> vec, T number) {
	auto itr = find(vec.begin(), vec.end(), number);
	size_t index = distance(vec.begin(), itr);
	if (index != vec.size()) return 1;
	else return 0;
}
ll frac(ll n){
	if (n == 0) return 1;
	return (n * frac(n - 1)) % MOD;
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// const int MAX_ = 1000000001;
// long long fac[MAX_], finv[MAX_], inv[MAX_];
//
// // テーブルを作る前処理
// void COMinit() {
//     fac[0] = fac[1] = 1;
//     finv[0] = finv[1] = 1;
//     inv[1] = 1;
//     for (int i = 2; i < MAX_; i++){
//         fac[i] = fac[i - 1] * i % MOD;
//         inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
//         finv[i] = finv[i - 1] * inv[i] % MOD;
//     }
// }
//
// // 二項係数計算
// long long COM(int n, int k){
//     if (n < k) return 0;
//     if (n < 0 || k < 0) return 0;
//     return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
// }
//

// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

ll cmod(ll n, ll k) {
  ll a = 1, b = 1;
  fi(1, k+1) {
    (a *= (n - i + 1)) %= MOD;
    (b *= i) %= MOD;
  }
  // return a / b;
  return a * modinv(b, MOD) % MOD;
}

int main(int argc, char** argv) {
#ifdef HIRO
  string quiz(argv[1]);
  string id(argv[2]);
  string filename("input/" + quiz + "/" + id + ".txt");
  ifstream fin(filename);
  if (fin.peek() == ifstream::traits_type::eof()) {
    cout << "No input.";
    return 0;
  }
  ifstream cin(filename);
  std::cin.rdbuf(cin.rdbuf());
#endif
  ll ans = 0;
  // ll n;
  // cin >> n;
  // ll n, m;
  // cin >> n >> m;
  ll n, m, k;
  cin >> n >> m >> k;
  
  // the sum of combination is 2^n -1
  // answer is 2^n - 1 - nCm - nCk
  // COMinit();

  // cout << cmod(200000, 199999) << endl;

  ans = modpow(2, n, MOD) - 1 - cmod(n, m) - cmod(n, k);

  while(ans < 0) ans += MOD;

  cout << ans << endl;
  return 0;
}



Measured memory_bytes: 256