 Solution

For a binary linear code $C\le {\mathbb{F}}_2^n$, its rank is its dimension of a vector space $r$. Use the weight enumerator conventionwhere $w(c)$ is the Hamming weight. This convention is essential here: $W_C(1,0)$ counts words of weight $n$, rather than zero words. There are $2^r$ codewords, so $\boxed{W_C(1,1)=2^r}$.

There is only one binary word of weight $n$, namely $1=(1,\dots ,1)$. Thus $W_C(1,0)=1$ precisely when $1\in C$. If this holds, the linear map $c\mapsto c+1$ is a bijection of $C$ and replaces weight $w$ by $n-w$; hence $W_C(s,t)=W_C(t,s)$. Conversely this symmetry gives $W_C(1,0)=W_C(0,1)=1$, because the zero word is the unique word of weight zero.

The repetition code consists of $0,1$, givingThe single parity-check code consists of all words of even weight. Exactly $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ words have weight $j$, so the binomial theorem gives