Problem:
 Score : 400 points    Problem Statement Akari has n kinds of flowers, one of each kind.  She is going to choose one or more of these flowers to make a bouquet.  However, she hates two numbers a and b , so the number of flowers in the bouquet cannot be a or b .  How many different bouquets are there that Akari can make?  Find the count modulo (10^9 + 7) .  Here, two bouquets are considered different when there is a flower that is used in one of the bouquets but not in the other bouquet.      Constraints  All values in input are integers.  2 \leq n \leq 10^9  1 \leq a < b \leq \textrm{min}(n, 2 \times 10^5)         Input Input is given from Standard Input in the following format:  n  a  b       Output Print the number of bouquets that Akari can make, modulo (10^9 + 7) . (If there are no such bouquets, print 0 .)        Sample Input 1 4 1 3      Sample Output 1 7  In this case, Akari can choose 2 or 4 flowers to make the bouquet.  There are 6 ways to choose 2 out of the 4 flowers, and 1 way to choose 4 , so there are a total of 7 different bouquets that Akari can make.       Sample Input 2 1000000000 141421 173205      Sample Output 2 34076506  Print the count modulo (10^9 + 7) .

Solution (Python):
import sys

stdin = sys.stdin

ns = lambda: stdin.readline().rstrip()
ni = lambda: int(stdin.readline().rstrip())
nm = lambda: map(int, stdin.readline().split())
nl = lambda: list(map(int, stdin.readline().split()))

n,a,b = nm()
mod = 10**9 + 7
s = pow(2,n,mod) - 1

def fur(n,r):
    p,q = 1,1
    for i in range(r):
        p = p*(n-i)%mod
        q = q*(i+1)%mod
    return p * pow(q,mod-2,mod) % mod

print((s - fur(n,a) - fur(n,b)) % mod)

Solution (Python):

from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    def comb(a,b):
        res = 1
        for i in range(b):
            res *= n-i
            res %= mod
            res *= pow(i+1,mod-2,mod)
            res %= mod
        return res
    n,a,b = LI()
    ans = pow(2,n,mod)-1
    ans -= comb(n,a)+comb(n,b)
    ans %= mod
    print(ans)
    return


if __name__ == "__main__":
    solve()

Solution (Python):
N,A,B=map(int,input().split())
MOD=10**9+7
def comb(n,k):
    if n<k:
        return 0
    if n<0 or k<0:
        return 0
    k=min(n-k,k)
    ans=1
    inv=[1]*(k+1)
    if k>=1:
        ans*=(n-k+1)%MOD
    for i in range(2,k+1):
        inv[i]=MOD-inv[MOD%i]*(MOD//i)%MOD
        ans=ans*(n-k+i)*inv[i]%MOD
    return ans
print((pow(2,N,MOD)-1-comb(N,A)-comb(N,B))%MOD)

Solution (Python):
import sys

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  

n, a, b = na()
mod = 1000000007

def comb(n, r, mod):
    if n < 0 or r < 0 or r > n: return 0
    nu = 1
    de = 1
    for i in range(r):
        nu = nu * (n-i) % mod
        de = de * (i+1) % mod
    return nu * inv(de, mod) % mod

def inv(a, mod):
    b = mod
    p, q = 1, 0
    while b > 0:
        c = a//b
        d = a; a = b; b = d % b
        d = p; p = q; q = d-c*q
    return p+mod if p < 0 else p

print((pow(2, n, mod) - comb(n, a, mod) - comb(n, b, mod) + 2*mod + mod - 1)%mod)