There exists an edge coloring of $K_{10,10}$ with two colors such that there are at most $56$ monochromatic copies of $K_{3,3}$. Show that there exists an edge coloring of $K_{10,10}$ with two colors such that there are at most $56$ monochromatic copies of $K_{3,3}$.

There are $100$ edges in $K_{10,10}$ and $9$ edges in $K_{3,3}$.
If we consider $56$ copies of $K_{3,3}$, then there are a total of   $56*9 = 504$ edges which is much more than $100$. So there is certainly intersections between the monochromatic copies of $K_{3,3}$.

I am not getting any idea how to approach the problem.
Any hints will be appreciable. Thanks.

Let $C=\{1,2\}$ be the colors and $K_{10,10}=(V_1\cup V_2,E)$. Let $\mathcal V_i=\{U\subseteq V_i:|U|=3\}$ be the $3$-subsets of $V_i$ for $i\in\{1,2\}$, and for $U_i\in V_i$ let $G(U_1,U_2)=(U_1\cup U_2,E(U_1,U_2))$ be the copy of $K_{3,3}$ with vertex sets $U_1$, $U_2$. For an assignment $x\in C^{E}$ to the colors and $c\in C$ let $S(x,U_1,U_2,c)=\{\forall e\in E(U_1,U_2)\,x(e)=c\}$ be the indicator that $G(U_1,U_2)$ is monochromatic of color $c$ under $x$. Then the set of monochromatic copies of $K_{3,3}$ in $K_{10,10}$ is $$M(x)=\{(U_1,U_2)\in\mathcal V_1\times\mathcal V_2:S(x,U_1,U_2,1)+S(x,U_1,U_2,2)=1\}.$$
Now, draw $X\in C^{E}$ uniformly at random. Then we have
$$\mathbb E[|M(X)|]=\sum_{U_1,U_2,c}\mathbb P(\forall e\in E(U_1,U_2)\,X(e)=c)
=\binom{10}{3}^2\cdot 2\cdot2^{-9}=\frac{225}{4}.$$
Markov's inequality yields $\mathbb P(|M(X)|\ge 57)\le\mathbb E[|M(X)|]/57=75/76<1$. This shows that there exists a coloring $x\in C^E$ such that $|M(x)|\le 56$, otherwise the probability would be $1$.
