Inserting random numbers from 1 to $n^2$ in a matrix of size $n \times n$ I have two matrices of size nxn with random numbers that are in range of $1$ to $n^2$.
I'm trying to calculate the probability of :


*

*the numbers 1 and 9 are present in the same indices in the two matrices,for example($n=4$):
\begin{bmatrix}
    14& 10 & 2  & 1 \\
    11& \textbf{1} & 16 & 15 \\
    4 & 9& 13 & \textbf{9}\\
    6 & 3& 5 & 7\\
\end{bmatrix}
\begin{bmatrix}
    13& 4& 2  & 7 \\
    15& \textbf{1} & 11& 10\\
    6 & 16& 5& \textbf{9}\\
    12 & 8& 3 & 14\\
\end{bmatrix}

*A pair (one or more) of two numbers that are sequential that exist in the same index of the two matrices.
for example(n=4):
\begin{bmatrix}
    14& 10 & 2  & 1 \\
    11& 1 & 16 & 15 \\
    4 & 9& 13 & \textbf{6}\\
    6 & 3& \textbf{5} & 7\\
\end{bmatrix}
\begin{bmatrix}
    13& 4& 2  & 7 \\
    15& 3 & 11& 10\\
    6 & 16& 2& \textbf{6}\\
    12 & 8& \textbf{5} & 14\\
\end{bmatrix}

*Row values of the first matrix are equal to the row values of the second matrix (not necessarily in the same index values) ,example(n=4):
\begin{bmatrix}
    14& 10 & 2  & 8 \\
    11& 1 & 16 & 15 \\
    4 & 12& 9& 6\\
    13 & 3& 5 & 7\\
\end{bmatrix}
\begin{bmatrix}
    10& 2& 8  & 14 \\
    16& 11 & 15& 1\\
    6 & 9& 4& 12\\
    13 & 3& 7 & 5\\
\end{bmatrix}

I'll solve the first part of the problem and leave the rest for you to solve
First place 1 and 9 at any two locations out of $n^2$ available locations. Then fill the rest of $2n^2-4$ with any number between 1 and $n^2$. So the total number of desired possibilities $=^{n^2}\!\!\!P_{2}(n^2)^{2n^2-4}$.
The number of possible matrix pairs without any restrictions is  $(n^2)^{2n^2}$.
Hence the probability is $\frac{n^2-1}{n^6}$
