<s>[INST] Number of ordered pairs of $A,B$ in Probability 
Let $|X|$ denote the number of elements in a set $X,$ 
Let $S = \{1,2,3,4,5,6\}$ be a sample space,
where each element is equally likely to occur. 
If $A$ and $B$ are independent events associated with
  $S,$ 
Then the number of ordered pairs $(A,B)$ 
such that $1 \leq |B| < |A|,$ equals

what i try
Let number of elements in $A,B,A \cap B$ is $x,y,z$ respectively
conditions given as $1\leq y\leq x.$
For $2$ Independent events $A,B$
is $\displaystyle P(A \cap B)=P(A)\cdot P(B)\Rightarrow \frac{z}{6}=\frac{x}{6}\cdot \frac{y}{6}\Rightarrow z=xy/6$
 [/INST] The condition should be $1\le y\lt x$, not $\le$.
Since $z=xy/6$, $x$ or $y$ must be divisible by $2$ and by $3$, and $xy\le6$. With $1\le y\lt x$, that allows only two solutions: $x=6$, $y=1$ and $x=3$, $y=2$.
In the first case, we have $6$ choices for the element in $y$.
In the second case we have $z=3\cdot2/6=1$, so there must be exactly one common element. That gives us $6$ choices for the common element, then $5$ choices for the other element in $y$ and then $\binom42=6$ choices for the other two elements in $x$.
Thus the total number of ordered pairs is $6+6\cdot5\cdot6=186$.
</s>