Graph is Bell Curve. Why?

We have aCb=a!/((a-b)!b!) we set a system of inequalities (a-(n-1)C n-1)<(a-nCn) and (a-nCn)>(a-(n+1)C(n+1)) first gives us(integers) unite(n<=276;n>724) and second 275<n<=723 we see that only 276 satisfies both

I love Haskell:

Prelude> let fact n = if (n <= 0) then 1 else n*(fact (n-1)) 

Prelude> (snd.maximum)[(div (fact (1000 - n)) ((fact(1000-2*n))*fact n), n) | n<-[1..500]]
276

Use stirling approximation and then differentiate. We get n =276.369

I used Alex Martelli's gmpy2 module on the grounds that he is a world-class programmer and would release only excellent code. Although the numbers are enormous this code runs very quickly.

java code:

public class brilliant201409061323{
    public static void main(String args[]){
        double max = 0;
        int n = 0;
        for(int i=0;i<=1000;i++){
            double c = C(1000-i,i);
            if(c>max){
                max = c;
                n = i;
            }
        }
        System.out.println(n+":"+max);
    }
    private static double C(int n,int r){
        double res = 1;
        for(double i=1;i<=r;i++){
            res *= n-i+1;
            res /= i;
        }
        return res;
    }
}

def factorial(n): m = 1 for i in range(1,n+1): m *= i return m

def combin(x,y): return(factorial(x)/(factorial(y)*factorial(x-y)))

coeffs = []

for n in range(1,1000): coeffs.append(combin(1000-n,n))

print(coeffs.index(max(coeffs)))

Sage code

print(max([(binomial(1000 - n, n), n) for n in xrange(1, 500)])[1])

returns

276

In Python, write:

def factorial(n):
product = 1
for i in range(n):
    product = product * (i+1)
return product

def combination(n,k):
combo = factorial(n)/(factorial(k) * factorial (n-k))
return combo

def binom_max(n):
binom_maximum = 0
k_max = 0
for j in range(n+1):
    comb_val = combination(n-j, j)
    if comb_val >= binom_maximum:
        binom_maximum = comb_val
        k_max = j
return k_max, binom_maximum

The maximum is achieved when the counter is at $j=276$ .