Parity does matter!

We will make use of a consequence of Lucas' Theorem , whereby a binomial coefficient $\binom{m}{n}$ is odd if and only if none of the digits in the binary expansion of $n$ is greater than the corresponding digit in the binary expansion of $m.$

In this case we have $m = 2015,$ which has a binary expansion of $11111011111.$ So all those binomial coefficients $\binom{2015}{n}$ in the expansion of $(a + b)^{2015}$ for which $n$ does not have a $1$ in the $6$ th from the right digit in their binary expansion will be odd. Since all numbers from $2016$ to $2047$ do have a $1$ in this position, we can then conclude that the number of values $n$ from $0$ to $2015$ inclusive that do not have a $1$ in this position will be $\frac{2048}{2} = \boxed{1024}.$

If writing out the Pascal's triangle, one can see that the number of odd binomial coefficients in row x = $2^k$ where k is the number of 1's in the binary form of the x. In this case, x = 2015, as the 2015th row of the Pascal triangle corresponds to the coefficients of the expansion of $(a+b)^{2015}$ . 2015 = 11111011111, There are 10 1's in this number, so the number of odd coefficients is $2^{10}$ = 1024.

Let's do it for a general case. That is, finding number of coefficients in the expansion of $(a+b)^n$ which are not divisible by $p$ . Here $p$ is a prime number. Theorem : If $n=\displaystyle\sum_{i=0}^{k}a_{i}p^{i}$ or in other words, $n=a_{k} a_{k-1} ... a_{2} a_{1} a_{0}$ in base $p$ . The number of coefficients not divisible by $p$ are $\displaystyle\prod_{i=0}^{k}(a_i+1)$ . Proof : For simplicity, let us assume that $a=1$ and $b=x$ . Thus we have $(1+x)^{n}=(1+x)^{\sum_{i=0}^{k}a_{i}p^{i}}=\displaystyle\prod_{i=0}^{k} \left[(1+x)^{p^i}\right]^{a_i}$ . If we divide the expression by $p$ , all the terms having coefficients divisible by $p$ will vanish. And we will be left with the coefficients that are not divisible by $p$ , which is desired. Now we will use the fact that $(1+x)^{p^i} \equiv (1+x^{p^i}) \text{ (mod }p)$ . So, $(1+x)^{n}\equiv \displaystyle\prod_{i=0}^{k} \left(1+x^{p^i}\right)^{a_i}$ Number of terms in the expansion of $\prod_{i=0}^{k} \left(1+x^{p^i}\right)^{a_i}$ is $(a_i+1)$ . Total number of terms will be $\displaystyle\prod_{i=0}^{k}(a_i+1)$ . Q.E.D. For the given question, $2015=11111011111_2$ , add 1 to all the digits, and multiply them. $22222122222$ thus number of odd terms will be $2^{10}=1024$ .