Find $\Pr(\textbf{R}^2=\mathbf{0})$ if $\textbf{R}_{4\times 4}$ is a matrix with 1 in 2 random positions and zeros otherwise 
Find $\Pr(\textbf{R}^2=\mathbf{0})$ if $\textbf{R}_{4\times 4}$ is a matrix with 1 in 2 random positions and zeros otherwise. 

(ITA entrance exam, Brazil, 2020)
My attempt: let $\textbf{R}=[r_{ij}]$ and $\textbf{R}^2=[r^2_{ij}]$, so that
$$r^2_{ij}=\sum_{t=1}^4 r_{it} r_{tj},\forall i,j\in\{1,..,4\}\ \ \ (*).$$
(1) there are ${16\choose 2}=120$ ways to position the 2 ones in the matrix. And it is direct to see that if at least one 1 is in diagonal, $\textbf{R}^2\not =\mathbf{0}$.
(2) let us count the number of $\textbf{R}$ matrices leading to $\textbf{R}^2\not =\mathbf{0}$ considering  3 cases:
Case 1 - Two 1s in the diagonal: ${4\choose 2}=6$ ways.
Case 2 - One 1 in the diagonal  and the other off diagonal: ${4\choose 1}{12\choose 1}=48$ ways.
Case 3 - Both ones off diagonal. Considering the notation on expression (*) above, let us consider $t=1$ for instance, so that $i$ and $j$ $\in \{2,3,4\}$, to avoid elements in the diagonal. There are $3^2=9$ possible ways to have $r_{i1}r_{1j}=1$ in this case. Considering $t\in \{2,3,4\}$, using the same argument, will lead to a a total of $4\times 3^2=36$ matrices $\textbf{R}$ with two 1s off diagonal leading to $\textbf{R}^2\not =\textbf{0}.$
By adding Cases 1 to 3, there are $6+48+36=90$ matrices $\textbf{R}$, leading to  $\textbf{R}^2\not =\textbf{0}$  or $120-90=30$ matrices $\textbf{R}$ with
$\textbf{R}^2 =\textbf{0}.$
And the asked probability would be $30/120$
But this last answer is wrong as the argument for case 3 overcounts, as the total should be 30 and not 36, with final correct answer $36/120=3/10$
Asked: Please point me out the mistake I'm doing the counting in Case 3. And advise on an efficient and straightforward way to count cases in situations like this.
 A: For case $3$, here is how I would personally approach it (not that your approach is worse, it's just a matter of what order I try things in): Let's place one off-diagonal $\color{red}1$ first. There are $12$ places to choose from. Say $r_{ij} = \color{red}1$. For each of those, there are $5$ off-diagonal spots that the second $\color{blue}1$ can be placed in, namely anywhere that is either in row $j$ or in column $i$. This makes $12\cdot 5 = 60$ total possibilities. However, we have double counted, as $r_{ij} = \color{red}{1}, r_{ab} = \color{blue}{1}$ is here counted separately from $r_{ab} = \color{red}{1}, r_{ij} = \color{blue}{1}$, so the final answer is $60\div 2 = 30$.
What you overcount in your breakdown is that each case of $i = j$ is counted twice. For instance, say $r_{12} = r_{21} = 1$. Then you count that occurrance once when $t = 1$, and once when $t = 2$. There are six such cases, which is exactly the six too many that you have.
(Note in my approach where I have $5$ instead of $6$, even though each row and each column have $3$ off-diagonal spots, and $3+3 = 6$. That's the exact same pitfall, namely when $i = j$, that I have taken care not to include more than once.)
As for the general tip, this is a good solution. You have just forgotten to look out for cases that are counted multiple times. Keep that in mind for next time, and you will be good to go, from what I've seen here.
A: There are ones in two places, $(i,j)$ and $(s,t)$. I will assume that these two ones are distinct, i.e. there is a first "one" and a second "one," so the probability space has $16\cdot 15=240$ outcomes, instead of $\binom{16}2=120$. This does not affect the resulting probability, since we double the size of the probability space and the number of favorable outcomes in the event.
There are two ways to have $R^2\neq 0$:


*

*If either $i=j$ or $s=t$, then $R^2$ will have a nonzero diagonal entry. The probabilibity of either of these occurring is $1$ minus the chance that both ones are off the diagonal, which is $1-\frac{12}{16}\frac{11}{15}$. 

*Both $i\neq j$ and $s\neq t$. In what follows, we will compute the conditional probability that $R^2\neq 0$ given the event $E$ that both ones are off-diagonal.
Conditional on $E$, the only way to have $R^2\neq 0$ for at least one of the two events to occur:


*

*$j=s$, in which case $(R^2)_{i,t}=1$ or 

*$i=t$, in which case $(R^2)_{s,j}=1$. 


Therefore,
\begin{align}
P(R^2\neq 0|E)
&=P(j=s\cup i=t|E)
\\&=P(j=s|E)+P(i=t|E)-P(j=s\cap i=t|E)
\\&=\frac{4\cdot 3\cdot 3}{12\cdot 11}+\frac{4\cdot 3\cdot 3}{12\cdot 11}-\frac{4\cdot 3}{12\cdot 11}=\frac{60}{12\cdot 11}
\end{align}
Finally, 
$$
\begin{align}
P(R^2\neq 0)
&=P(R^2\neq 0|E^c)P(E^c)+P(R^2\neq 0|E)P(E)
\\&=1\cdot (1-\tfrac{12}{16}\tfrac{11}{15})+\tfrac{60}{12\cdot 11}\tfrac{12\cdot 11}{16\cdot15}
\\&=\frac{3}{10}.
\end{align}
$$
A: $R^2$ is not necessarily zero when both $1$s appear at symmetric off-diagonal positions. E.g.
$$
\pmatrix{0&1&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0}^2=\pmatrix{1&0&0&0\\ 0&1&0&0\\ 0&0&0&0\\ 0&0&0&0}.
$$
