Maximize the year!

If you find the binomial coefficients $\dbinom{n}{r}$ by using Pascal's triangle you find that for each $n$ and $r$ ranging from 1 to $n$ , the row is symmetrical, rising steadily to a maximum and then steadily decreasing.

So the maximum value of $\dbinom{2015}{r}$ occurs in the middle of the $2015^{th}$ row of the triangle, in other words when $r=1007\text{ or }r=1008$

So the minimum value of $r$ is $\boxed{1007}$

What we have to show is that $\dbinom{2n+1}{n} > \dbinom{2n+1}{n-a} \forall a \in N$ Now by comparing terms in the following two products we see that: $(n-a+1)(n-a+2) \cdots n < (n+2)(n+3) \cdots (n+a)(n+a+1)$ $\implies n!(n+1)! < (n-a)!(n+a+1)!$ $\Rightarrow\frac{(2n+1)!}{(n-a)!(n+1+a)!} < \frac{(2n+1)!}{n! (n+1)!}$ $\therefore\dbinom{2n+1}{n} > \dbinom{2n+1}{n-a} \forall a \in N$

(n,r) = n! / (n! * (n-r)! ) denominator should be minimum for the expression value to be maximum

n! * (n-r)! would be minimum when r = n/2 when n is even when n is odd, the minimum is attained when r = (n-1)/2 or when r = (n+1)/2

hence for our case, the minimum value of the two is (2015-1)/2 = 1007

You may use Pascal's Triangle.