A probability problem by Vijay Bhava

A corollary of Lucas' Theorem is that $\binom{n}{r}$ is odd if and only if each bit in the binary expansion of $r$ is less than or equal to the corresponding bit in the binary expansion of $n$ .

As the binary expansion of $71$ is $1000111$ , for $\binom{n}{r}$ to be odd the binary expansion of $r$ can have either a $0$ or $1$ in the $2^{6}, 2^{2}, 2^{1}, 2^{0}$ bits but must have a $0$ in the $2^{5}, 2^{4}, 2^{3}$ bits. This results in $2^{4} = \boxed{16}$ integral values of $r$ for which $\binom{71}{r}$ is odd.

Comments: In general, if there are $m$ $1$ 's in the binary expansion of $n$ then $2^{m}$ of the binomial coefficients $\binom{n}{k}$ will be odd. If $n = 2^{q} - 1$ for some positive integer $q$ then all $n + 1$ of the binomial coefficients will be odd.