Let $n$ be a positive integer and $k$ be the number of $1$s in $n$'s binary representation. Show there are $2^k$ odd integers in $\binom{n}{0}$, $\binom{n}{1}$, $\cdots$, $\binom{n}{n}$.
Let the binary representation of $n$ be $b_tb_{t-1}\cdots b_1b_0$. Because $k$ is the number of $1$s in $n$'s binary representation, $n$ can be written as $$n=2^{i_1} + 2^{i_2} + \cdots + 2^{i_k}$$
where $i_1 > i_2 > \cdots > i_k$ are all non-negative integers.
Then, using the reasoning presented in # 4305, we find odd-even parity of $\binom{n}{k}$, where $k=0$, $1$, $\cdots$, $n$, depends on whether $k$ can be represented as $$k=c_1\cdot 2^{i_1} + c_2\cdot 2^{i_2} + \cdots + c_k\cdot 2^{i_k}$$
where $c_1$, $c_2$, $\cdots$, $c_k$ are either $0$ or $1$. When such a representation exists, $\binom{n}{k}$ will be odd. Otherwise, it will be even.
Because, each $c_i$ has two choices, and there are $k$ such $c_i$, therefore there are totally $2^k$ numbers can be written in this way. This means there are totally $2^k$ odd integers.