Combinatorics problem about odd counts There are boys and girls studying at a school. A group of boys is called good if every girl knows at least one boy from the group. A group of girls is called good if every boy knows at least one student in the group. Prove that if the number good boy groups is odd, the number of girl good groups is also odd. (knowledge is mutual) 
The problem is from a Romanian National Olympiad, and I tried solving it for about two hours but made no apperant progress. 
I will be happy if someone can explain it to me. Thank you in advance! 
 A: Let $B$ denote the set of boys and let $G$ denote the set of girls.
Let $R\subseteq B\times G$ denote the (symmetric) relation of knowing
eachother.
For $g\in G$ let $B_{g}=\left\{ b\in B\mid\neg bRg\right\} $ and
for $K\subseteq G$ let $B_{K}=\bigcap_{g\in K}B_{g}$.
Then a group of boys $H\subseteq B$ is bad if and only if $H\in\wp\left(B_{g}\right)$
for some $g\in G$ and with inclusion/exclusion we find the cardinality
of the collection of these groups as the following expression:
$$\left|\bigcup_{g\in G}\wp\left(B_{g}\right)\right|=\sum_{\varnothing\neq K\subseteq G}\left(-1\right)^{^{\left|K\right|-1}}\left|\cap_{g\in K}\wp\left(B_{g}\right)\right|=\sum_{\varnothing\neq K\subseteq G}\left(-1\right)^{^{\left|K\right|-1}}\left|\wp\left(\cap_{g\in K}B_{g}\right)\right|=$$$$\sum_{\varnothing\neq K\subseteq G}\left(-1\right)^{^{\left|K\right|-1}}\left|\wp\left(B_{K}\right)\right|\tag1$$
We are not interested in the cardinality itself but in the parity
of this cardinality. 
Further this parity is the same as the parity
of the cardinality of the collection of good groups of boys because
addition of both cardinalities must give the even number $2^{\left|B\right|}$. 
Calculating modulo $2$ we find that only the terms in $(1)$ where $B_{K}=\varnothing$
matter, because the others are all even numbers. Also the powers of $-1$ are irrelevant by calculation modulo $2$.
This leads to the conclusion that the parity of the cardinality of the set $\left\{ K\in\wp\left(G\right)-\left\{ \varnothing\right\} \mid B_{K}=\varnothing\right\} $
must be the same.
Now observe that:
$$B_{K}=\varnothing\iff\neg\exists b\in B\forall g\in K\;\neg bRg\iff\forall b\in B\exists g\in K\;bRg\iff K\text{ is a good girl group}$$
This completes the proof.
