Let $\left| X \right|$ 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 event associated with $S$ , then the number of ordered pair $\left( A,B \right)$ such that $1\le \left| B \right|<\left| A \right|$ equal are

Hint: We are given that $A$ and $B$ are independent events. We will start our solution by considering that $A$ has $a$ number of elements while $B$ has $b$ number of elements. Then we will use the formula $P\left( A \right)\times P\left( B \right)=P\left( A\cap B \right)$ . We will find the probability of $A,\text{ }B\text{ }and\text{ }A\cap B$. We will get a condition as $ab=6c$, then we will consider cases of possible value of $a$ and $b$ by taking values of c from 1 to 5. We will keep in mind while considering cases that $\left| B \right|<\left| A \right|\text{ }and\text{ }1\le \left| B \right|$ and using this we find an ordered pair possible.

Complete step-by-step answer:
We have been given a sample space $S$ as $S=\{1,2,3,4,5,6\}$ . So we can write the number of elements as $\left| S \right|=6$ .
We have also been given that $A$ and $B$ are two independent events. Let us assume $A$ consist of $a$ element while $B$ consist of $b$ element.
Now since $A$ and $B$ are independent, so we can write that
 $P\left( A \right)\times P\left( B \right)=P\left( A\cap B \right)...............\left( 1 \right)$
Let us assume the number of elements in $A\cap B$ is $c$ .
Now as total outcome possible are 6 in number as we have $\left| S \right|=6$ , so we can write the probability as
$\begin{align}
  & P\left( A \right)=\dfrac{a}{6} \\
 & P\left( B \right)=\dfrac{b}{6}\text{ }and \\
 & P\left( A\cap B \right)=\dfrac{c}{6} \\
\end{align}$
Now using this in $\left( 1 \right)$ , we get
 $\begin{align}
  & P\left( A \right)\times P\left( B \right)=P\left( A\cap B \right) \\
 & \Rightarrow \dfrac{a}{6}\times \dfrac{b}{6}=\dfrac{c}{6} \\
\end{align}$
Simplifying we get
 $ab=6c............\left( 2 \right)$
Now we are asked to find the number of ordered pairs of $\left( A,B \right)$ . To find the ordered pair we should know the possible number of elements $A$ and $B$ can have.
Now we use $eq\left( 2 \right)\text{ }ab=6c$ to find our possible value of $a\text{ }and\text{ }b$ .
$Case\text{ }I,$ if $c=1$
If we put $c=1$ in $ab=6c$ , we get $ab=6$ .
So we have one possibility as \[\left. i \right)a=6,b=1\].
So using concept of combinations, we can write the number of possible ordered pairs as
 $\Rightarrow {}^{6}{{C}_{6}}\times {}^{6}{{C}_{1}}=6$
Next possibility is \[\left. ii \right)a=3,b=2\] .
So using concept of combinations, we can write the number of possible ordered pairs as $\Rightarrow {}^{6}{{C}_{3}}\times {}^{3}{{C}_{1}}\times {}^{3}{{C}_{1}}=180$
As $\left| A \right|>\left| B \right|$ so \[\left( iii \right)a=2,\text{ }b=3\text{ }and\text{ }\left( iv \right)a=1,\text{ }b=6\] are not possible.
Now, let us take $Case\text{ }II,$ if $c=2$ .
Then putting $c=2$ in $ab=6c$, we get $ab=12$ .
So one possibility we have is \[\left. i \right)a=6,b=2\] .
So, again using combinations, we get number of possible ordered pairs as
 $\Rightarrow {}^{6}{{C}_{6}}\times {}^{6}{{C}_{2}}=15$
Next, we have a possibility of \[\left. ii \right)a=4,b=3\]
So, again using combinations, we get number of possible ordered pairs as
$\Rightarrow {}^{6}{{C}_{4}}\times {}^{3}{{C}_{2}}\times {}^{2}{{C}_{1}}=180$
Again options like \[\left( iii \right)a=3,\text{ }b=4\text{ }and\text{ }\left( iv \right)a=2,\text{ }b=6\] are not possible as we have condition $\left| A \right|>\left| B \right|$.
Next, we will take $Case\text{ }III,$ if $c=3$
If $c=3$ then we get, $ab=18$ .
One of the possibilities is \[\left. i \right)a=6,b=3\]
Now, using combinations, we get number of possible ordered pairs as
 $\Rightarrow {}^{6}{{C}_{6}}\times {}^{6}{{C}_{3}}=20$
Next, we have $Case\text{ }IV,$ if $c=4$
If we use $c=4$ in $ab=6c$, we get $ab=24$
So, we have possibility as \[\left. i \right)a=6,b=4\]
So, again using combinations, we get number of possible ordered pairs as
\[\Rightarrow {}^{6}{{C}_{6}}\times {}^{6}{{C}_{4}}=15\]
Now, let us take $Case\text{ }V,$ if $c=5$
If we use $c=5$ in $ab=6c$ , we get $ab=30$.
So we have \[\left. i \right)a=6,b=5\]
So, using combinations, we get number of possible ordered pairs as
 $\Rightarrow {}^{6}{{C}_{6}}\times {}^{6}{{C}_{5}}=6$
Now as our event $A$ and $B$ are independent so are the total ordered pairs in the sum of all these possible ordered pairs. So we get
 $\begin{align}
  & \Rightarrow 6+180+180+15+20+15+6 \\
 & =422 \\
\end{align}$
So we get the total number of ordered pairs $=422$ .

Note: Students must note here that the case when $c=6$ is not mentioned because $c$ represents the number of elements that lie in $A\cap B$ . If 6 elements belong to $A\cap B$ it would mean that $A$ and $B$ are exactly same, but we are strictly given that $\left| A \right|>\left| B \right|$ , meaning that $A$ is a greater set than $B$ and that’s why the case when $c=6$ is not mentioned.