n, a, b = map(int, input().split())
mod = 10 ** 9 + 7

#n種類の花を使って、花束を作る組み合わせは2^n-1通り
#pow(x,y,mod) : (x**y) % mod
ans = pow(2, n, mod)-1

#nCa
nCa = 1
#分子:n*(n-1)*...*(n-a+1)
for i in range(n-a+1, n+1):
    nCa *= i
    nCa %= mod
#分母:1/a! = 1/(1*2*...*a) = 1^(p-2)*2^(p-2)*,,,*a^(p-2)
for i in range(1, a+1):
    nCa *= pow(i, mod-2, mod)
    nCa %= mod
    
#nCb
nCb = 1
for i in range(n-b+1, n+1):
    nCb *= i
    nCb %= mod
for i in range(1, b+1):
    nCb *= pow(i, mod-2, mod)
    nCb %= mod

#2^n-1通りからnCaとnCbを引く
ans -= (nCa + nCb)
ans %= mod
print(ans) n, a, b = map(int, input().split()) n map(int, input().split()) map int input().split() ().split () input split a b mod = 10 ** 9 + 7 mod 10 ** 9 + 7 10 ** 9 10 9 7 ans = pow(2, n, mod)-1 ans pow(2, n, mod)-1 pow(2, n, mod) pow 2 n mod 1 nCa = 1 nCa 1 for i in range(n-a+1, n+1):
    nCa *= i
    nCa %= mod
#分母:1/a! = 1/(1*2*...*a) = 1^(p-2)*2^(p-2)*,,,*a^(p-2) i range(n-a+1, n+1) range n-a+1 n-a n a 1 n+1 n 1 nCa *= i nCa i nCa %= mod nCa mod for i in range(1, a+1):
    nCa *= pow(i, mod-2, mod)
    nCa %= mod
    
#nCb i range(1, a+1) range 1 a+1 a 1 nCa *= pow(i, mod-2, mod) nCa pow(i, mod-2, mod) pow i mod-2 mod 2 mod nCa %= mod nCa mod nCb = 1 nCb 1 for i in range(n-b+1, n+1):
    nCb *= i
    nCb %= mod i range(n-b+1, n+1) range n-b+1 n-b n b 1 n+1 n 1 nCb *= i nCb i nCb %= mod nCb mod for i in range(1, b+1):
    nCb *= pow(i, mod-2, mod)
    nCb %= mod

#2^n-1通りからnCaとnCbを引く i range(1, b+1) range 1 b+1 b 1 nCb *= pow(i, mod-2, mod) nCb pow(i, mod-2, mod) pow i mod-2 mod 2 mod nCb %= mod nCb mod ans -= (nCa + nCb) ans nCa + nCb nCa nCb ans %= mod ans mod print(ans) print ans nCa *= i i nCa a, b = map(int, input().split()) map(int, input().split()) a mod = 10 ** 9 + 7 10 ** 9 + 7 mod b = map(int, input().split()) map(int, input().split()) b nCa = 1 1 nCa nCb = 1 1 nCb nCb %= mod mod nCb nCa %= mod mod nCa nCb *= pow(i, mod-2, mod) pow(i, mod-2, mod) nCb ans = pow(2, n, mod)-1 pow(2, n, mod)-1 ans nCa %= mod mod nCa n, a, b = map(int, input().split()) map(int, input().split()) n ans %= mod mod ans nCa *= pow(i, mod-2, mod) pow(i, mod-2, mod) nCa nCb %= mod mod nCb ans -= (nCa + nCb) nCa + nCb ans nCb *= i i nCb
n, a, b = map(int, input().split())
mod = 10 ** 9 + 7

#n種類の花を使って、花束を作る組み合わせは2^n-1通り
#pow(x,y,mod) : (x**y) % mod
ans = pow(2, n, mod)-1

#nCa
nCa = 1
#分子:n*(n-1)*...*(n-a+1)
for i in range(n-a+1, n+1):
    nCa *= i
    nCa %= mod
#分母:1/a! = 1/(1*2*...*a) = 1^(p-2)*2^(p-2)*,,,*a^(p-2)
for i in range(1, a+1):
    nCa *= pow(i, mod-2, mod)
    nCa %= mod
    
#nCb
nCb = 1
for i in range(n-b+1, n+1):
    nCb *= i
    nCb %= mod
for i in range(1, b+1):
    nCb *= pow(i, mod-2, mod)
    nCb %= mod

#2^n-1通りからnCaとnCbを引く
ans -= (nCa + nCb)
ans %= mod
print(ans)
