You know combinations: for example, 
if you take 5 cards from a 52 cards deck you have 2,598,960 different combinations.

In mathematics the number of x combinations you can take from a set of n elements
is called the binomial coefficient of n and x, or more often `n choose x`.
The formula to compute `m = n choose x` is: `m = n! / (x! * (n - x)!)`
where ! is the factorial operator.

You are a renowned poster designer and painter. You are asked to provide 6 posters 
all having the same design each in 2 colors. Posters must all have a different color combination and you have the choice of 4 colors: red, blue, yellow, green. 
How many colors can you choose for each poster?

The answer is two since `4 choose 2 = 6`. The combinations will be:
{red, blue}, {red, yellow}, {red, green}, {blue, yellow}, {blue, green}, {yellow, green}.

Now same question but you have 35 posters to provide and 7 colors available. How many colors for each poster?
If you take combinations `7 choose 2` you will get 21 with the above formula.
But 21 schemes aren't enough for 35 posters. If you take `7 choose 5` combinations you will get 21 too.
Fortunately if you take `7 choose 3` or `7 choose 4` combinations you get 35 and so each poster will have a different combination of
3 colors or 5 colors. You will take 3 colors because it's less expensive.

Hence the problem is: 

knowing `m` (number of posters to design), 
knowing `n` (total number of available colors), 
let us 
search `x` (number of colors for each poster so that each poster has a unique combination of colors and the number of combinations is exactly the same as the number of posters).

In other words we must find **x** such as `n choose x = m (1)` for a given m and a given n;
`m >= 0 and n > 0`. If many x are solutions give as result the smallest x.
It can happen that when m is given at random there are no x satisfying `equation (1)` then
return -1.

Examples:
```
checkchoose(6, 4) --> 2
checkchoose(4, 4) --> 1
checkchoose(4, 2) --> -1
checkchoose(35, 7) --> 3
checkchoose(36, 7) --> -1

a = 47129212243960
checkchoose(a, 50) --> 20
checkchoose(a + 1, 50) --> -1
```

Difficulty: introductory

Starter code:

def checkchoose(m, n):
	

Solutions:
def checkchoose(m, n):
    c = 1
    for x in range(n // 2 + 1):
        if c == m: return x
        c = c * (n-x) // (x+1)
    else: return -1


from math import factorial

def choose(n, x):
    return factorial(n) // (factorial(x) * factorial(n-x)) if x else 1

def checkchoose(m, n):
    for x in range(n+1):
        if choose(n, x) == m:
            return x
    return -1

from math import factorial as fact

def checkchoose(m, n):
    for k in range(0, (n)//2 + 1):
        if fact(n) // (fact(n-k) * fact(k)) == m:
            return k
    return -1

from math import factorial as fact

def combinations(n, k):
    return fact(n) // (fact(k) * fact(n-k))

def checkchoose(posters, colors):
    return min([k for k in range(colors//2 +1) if combinations(colors, k) == posters], default=-1)

def checkchoose(m, n):
    from math import factorial as f
    a = 0
    while a <= n//2:
        if m == f(n) / (f(a)*f(n-a)):
            return(a)
        else:
            a += 1
    return(-1)

from math import factorial as fact
def checkchoose(m, n):
    l = [i for i in range(1, n+1) if fact(n)/(fact(i)*fact(n-i))==m]
    return 0 if m<n else l[0] if l else -1

def checkchoose(m, n):
  for k in range(n + 1):
    nk = n - k
    ans = partial_fact(n, k) / factorial(nk) if k > nk else partial_fact(n, nk) / factorial(k)
    if m == ans: return k
    if m < ans: return -1 #No need to search the other side of the triangle
  return -1

f = [0, 1, 2, 6, 24, 120]
def factorial(n):
    if n == 0 or n == 1: return 1
    if n < len(f): return f[n]
    f.append(factorial(n - 1) * n)
    return f[n]
partial_fact = lambda hi, le: 1 if hi <= le else hi * partial_fact(hi - 1, le)

from operator import mul
from functools import reduce

def choose(n, p):
    if (p > n): 
        return 0
    if (p > n - p): 
        p = n - p
    return reduce(mul, range((n-p+1), n+1), 1) // reduce( mul, range(1,p+1), 1)

def checkchoose(m, n):
    mx = choose(n, n // 2)
    if (m > mx): 
        return -1
    i = 0
    while (i <= (n // 2) + 1):
        if choose(n, i) == m:
            return i
        i += 1
    return -1