Animated Solution for Mathematics - Permutations and Combinations: Let $X=\{1,2,3,4,5\}$. The number of different ordered pairs (Y,Z) that can formed such that $Y\subseteq X,Z\subseteq X$ and $Y\cap Z$ is empty is:

Have you ever looked at a set theory problem and felt like you were drowning in a sea of possible subsets? It is a common feeling.

When we see $Y\subseteq X$ and $Z\subseteq X$, our instinct is often to try and list all possible subsets of $X$, which is $2^5=32$, and then try to pair them up. But that path is a labyrinth of complexity.

Today, we are going to take a step back and look at this problem through the eyes of the elements themselves. This is the 'element-wise' perspective, and it is one of the most powerful tools in your JEE toolkit.

We start with our universal set $X=\{1,2,3,4,5\}$. We are tasked with forming ordered pairs $(Y,Z)$ such that $Y$ and $Z$ are subsets of $X$ and, most importantly, $Y\cap Z=\varnothing$.

This condition is our North Star. It tells us that $Y$ and $Z$ are disjoint.

In the language of Venn diagrams, if you draw a circle for $Y$ and a circle for $Z$ inside the box representing $X$, they will never touch. They are islands in the sea of $X$.

Instead of asking "What are the possible sets $Y$ and $Z$?", let us ask a simpler question: "What happens to the number 1?"

Imagine the number 1 is a traveler. It looks at the Venn diagram and sees three distinct regions where it can reside:

1. Region 1: Inside the circle of $Y$. If 1 goes here, it is in $Y$ and definitely not in $Z$.

2. Region 2: Inside the circle of $Z$. If 1 goes here, it is in $Z$ and definitely not in $Y$.

3. Region 3: Outside both circles. It is still in the universal set $X$, but it has chosen to join neither $Y$ nor $Z$.

Notice that there is no "Region 4" where 1 is in both $Y$ and $Z$, because the disjoint condition forbids it. So, for the number 1, there are exactly 3 choices.

Now, here is the beauty of this approach. Does the choice of the number 1 affect the choice of the number 2? Not at all.

The number 2 is also a traveler, and it looks at the same three regions. It can go to $Y$, $Z$, or neither, regardless of where 1 went. The same applies to 3, 4, and 5.

Each element is an independent decision-maker. We have 5 elements, and each has 3 independent choices. By the Fundamental Principle of Counting, the total number of ways to form these pairs is the product of the choices for each element:

When you calculate $3^5$, you get 243.

But the number itself is less important than the logic that got us there. We transformed a problem about sets into a problem about individual elements.

Whenever you face a constraint like $Y\cap Z=\varnothing$, stop and ask yourself: "How many regions can an element occupy?" Once you identify those regions, the problem collapses into a simple multiplication. Keep this 'element-wise' perspective in your arsenal—it is the key to unlocking many complex combinatorial problems in your JEE journey.