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.
 A: You can first simplify the problem:
Let 1 denote an odd number and 0 and even number.
Case 1: P(X=0)
$11111$
$00000$
The even number can be arranged in 5! ways. The odd numbers can be arranged in 5! ways, too. Thus you have the factor $(5!)^2$ in all cases. In each coloumn 0 can be in the first row or in the second row. This gives an additional factor of $2^5$. You have calculated P(X=0) right.
Case 2: P(X=1)
$00001$
$01111$
The columns can be arranged in $\frac{5!}{3!\cdot 1!\cdot 1!}$ ways.  There are 3 columns with different numbers. Thus the additional factor is $2^3$
Case 3: P(X=2)
$00011$
$00111$
The columns can be arranged in $\frac{5!}{2!\cdot 2!\cdot 1!}$ ways. There are 2x2 columns with equal numbers and one column with different numbers. Thus the additional factor is $2^1$
In total you get the distribution of X:
$\begin{array}{|c|c|c|c|} \hline  x & 0 & 1 & 2  \\ \hline  P(X=x) & \frac{5!\cdot 5!}{10!}\cdot 2^5  & \frac{5!\cdot 5!}{10!}\cdot \frac{5! \cdot 2^3}{3!} & \frac{5! \cdot 5!}{10!}\cdot \frac{2\cdot 5!}{2!\cdot 2!\cdot 1!} \\ \hline  \end{array}$
A: 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. 
