Probability distribution of number of columns that has two even numbers in a chart 
We distribute numbers $\{1,2,...,10\}$ in random to the following chart:

Let $X$ be the number of columns that has two even numbers.
What is the distribution of $X$?

My attempt:
$|\Omega|=10!$
$P(x=0)=\frac {(5!)^2\cdot 2 ^5}{10!}=\frac 8 {63}$ Explanation: Arrange the even and odd numbers, choose which one is up 5 times.
I'm having trouble with the rest, for $P(x=1)$: choose two even numbers and a column $5\binom  5 2$, this force a column with odd numbers so choose them and a column $4\binom  5 2$, now we're left with arranging the rest like before: $(3!)^2\cdot 2^3$ so:
$P(x=1)=\frac{5\binom  5 24\binom  5 2 (3!)^2\cdot 2^3}{10!}= \frac {10}{63}$ which is too small to be right, and if I use the same reasoning $P(x=2)$ is too small again so they don't add up to 1.

Let us grind it out. 
There are five odd and five even. So we might as well assume that we use five $0$'s and five $1$'s. Let $X$ be the number of columns with two $0$'s. We want the probability distribution of $X$. 
The random variable $X$ can only take on the values $0$, $1$, and $2$. So we really have only two probabilities to compute, since if we know two of $\Pr(X=0)$, $\Pr(X=1)$, and $\Pr(X=2)$ then the third is easy.
What is the probability that $X=2$? For that to happen, we need $2$ or $3$ $0$'s in the top row. Let us deal with $2$ $0$'s and then, because of symmetry, multiply by $2$. 
There are $\binom{10}{5}$ equally likely ways to place the $0$'s. Of these ways, $\binom{5}{2}\binom{5}{3}$ have $2$ $0$'s in the top row. For the locations in the top row can be chosen in $\binom{5}{2}$ ways, and for each of these ways the locations in the bottom row can be chosen in $\binom{5}{3}$ ways. So the probability of $2$ $0$'s in the top row is $\frac{\binom{5}{2}\binom{5}{3}}{\binom{10}{5}}$. 
Given there are $2$ $0$'s in the top row, what is the probability of $2$ columns of $0$'s? Without loss of generality we may assume the $0$'s in the top row occupy the first $2$ positions. We want the probability that $2$  of the bottom $0$'s are in the first $2$ positions. Of the $\binom{5}{3}$ ways that the bottom $0$'s can be arranged, $3$ have the desired property. So 
$$\Pr(X=2)=2\cdot\frac{\binom{5}{2}\binom{5}{3}}{\binom{10}{5}}\cdot \frac{3}{\binom{5}{3}}.$$
It's your turn! I suggest going after $\Pr(X=0)$. This can happen in two types of ways: all the $0$'s are in one of the rows or it is a 4-1 (or 1-4) distribution and there is no match. 
