#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vvvi = vector<vector<vector<int>>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;
using vs = vector<string>;
using vb = vector<bool>;
#define FOR(i, m, n) for (int i = (m); i < (n); i++)
#define FORR(i, m, n) for (int i = (m); i >= (n); i--)
#define REP(i, n) FOR(i, 0, (n))
#define REPR(i, n) FORR(i, (n) - 1, 0)
#define REP1(i, n) FOR(i, 1, (n) + 1)
#define REPS(c, s) for (char c : s)
#define ALL(c) (c).begin(), (c).end()
#define SORT(c) sort(ALL(c))
#define REV(c) reverse(ALL(c))
#define sz(v) (int)v.size()
#define endl '\n'
template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
template<class T> inline void prn(vector<T>& v) {int n = sz(v); REP(i, n) cout << v[i] << ' ';}
template<class T> inline void printv(vector<T>& v) {int n = sz(v); REP(i, n) cout << v[i] << (i == n - 1 ? endl : ' ');}
template<class T> inline void printvv(vector<vector<T>>& v) {for (auto u : v) printv(u);}
template<class T> inline void printlnv(vector<T>& v) {int n = sz(v); REP(i, n) cout << v[i] << endl;}
const int MOD = 1000000007;
const int INF = 1000000001;
const ll LINF = 1000000001000000001LL;
 
void solve();
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(numeric_limits<double>::max_digits10);
    solve();
    return 0;
}

vector<ll> list_mod_inverse(ll n, ll m) {
    vector<ll> inv(n + 1);
    inv[1] = 1;
    for (int i = 2; i <= n; ++i)
        inv[i] = inv[m % i] * (m - m / i) % m;
    return inv;
}

ll nck(ll n, ll k, ll m, vector<ll>& inv) {
    if (n - k < k) k = n - k;
    ll ret = 1;
    REP1(i, k) {
        ret *= ((n - i + 1) * inv[i]) % m;
        ret %= m;
    }
    return ret;
}

class mint {
        ll value;
        int mod;
    public:
        constexpr mint(ll src = 0, int m = 1000000007) noexcept : value((src % m + m) % m), mod(m) {}

        constexpr ll v() { return value; }

        constexpr mint operator+(const mint rhs) const noexcept {return mint(*this) += rhs;}
        constexpr mint operator-(const mint rhs) const noexcept {return mint(*this) -= rhs;}
        constexpr mint operator*(const mint rhs) const noexcept {return mint(*this) *= rhs;}
        constexpr mint operator/(const mint rhs) const noexcept {return mint(*this) /= rhs;}

        constexpr mint &operator+=(const mint rhs) noexcept {
            value += rhs.value;
            if (value >= mod) value -= mod;
            return *this;
        }

        constexpr mint &operator-=(const mint rhs) noexcept {
            if (value < rhs.value) value += mod;
            value -= rhs.value;
            return *this;
        }

        constexpr mint &operator*=(const mint rhs) noexcept {
            value = value * rhs.value % mod;
            return *this;
        }
        
        constexpr mint &operator/=(mint rhs) noexcept {
            ll exp = mod - 2;
            while (exp) {
                if (exp % 2) *this *= rhs;
                rhs *= rhs;
                exp /= 2;
            }
            return *this;
        }
        
        constexpr mint pow(ll p) noexcept {
            if (p == 0) return 1;
            else if (p % 2 == 1) return pow(p - 1) * *this;
            else { mint a = pow(p / 2); return a * a; }
        }
};

void solve() {
    ll n, a, b;
    cin >> n >> a >> b;
    auto inv = list_mod_inverse(200200, MOD);
    cout << (mint(2).pow(n).v() - 1 - nck(n, a, MOD, inv) - nck(n, b, MOD, inv) + MOD + MOD) % MOD << endl;
}


Measured memory_bytes: 4780