I have 9 entries $a_1,a_2,a_3,a_4,\cdots,a_9$ to place in matrix $G$ with equal probabilities. $$ G = \begin{bmatrix} \times & \times & \times \\ \times & \times & \times \\ \times & \times & \times \\ %0 & 0 & 0 & \dots & 0 & 1 & -g_{0,1}\lambda_1 \end{bmatrix}. $$ If $a_1,a_2,a_3$ are to be in the same row or the same column, I may calculate the probability of this even as: $(1\times \frac{2}{8} \times \frac{1}{7})\times 2=\frac{1}{14}$. I understand this as when I fix $a_1$, I have 2 options out of 8 options for $a_2$, and then I know both $a_1$ and $a_2$, I have only 1 option out of 7 options for $a_3$. The same argument is for the column. I hope this understanding is correct.
If $a_1$ and two of $a_2,a_3,a_4$ are to be in the same row or the same column (e.g., $a_1,a_3,a_4$; or $a_1,a_2,a_4$, etc...), I could not calculate that probability. Likewise, the probability for $a_1$ and two of $a_2,a_3,a_4,a_5$ are to be in the same row or the same column
Can someone please guide me?
Actually I need to expand this for more general $M\times N$ matrix. However, I am trying to understand a small matrix.
To see if $a_1$ shares a row (or column) with any two of $a_2$, $a_3$, or $a_4$, we calculate as follows:
Fix $a_1$ to be somewhere. Let's assume for convenience that it is located in the top left. Now, let's check to see if any two of $a_2$, $a_3$, or $a_4$ are in the same column. For the middle left value, we have a probability of $\frac{3}{8}$ that one of $a_2$, $a_3$, or $a_4$ fills that spot. After that, we have a probability of $\frac{2}{7}$ that one of the other two values fills the bottom left spot. Therefore, the probability that $a_1$ shares a column with any two of $a_2$, $a_3$, or $a_4$ is $1 \times \frac{3}{8} \times \frac{2}{7} = \frac{3}{28}$.
Now let's look at a general example:
Suppose you have an $M \times N$ matrix and you want to see if $a_1$ shares the same column with any of $n$ possible values. There are $(M-1)$ remaining spots to fill. If you've done combinations, you see that there are ${{MN-1}\choose{M-1}}$ ways to fill the remaining $(M-1)$ spots. There are ${{n}\choose{M-1}}$ ways to fill the remaining $(M-1)$ spots with any of the desired $n$ values. Then, the probability of filling the spots with our desired values will simply be:
$$\frac{{{MN-1}\choose{M-1}}}{{{n}\choose{M-1}}}$$