Animated Solution for Mathematics - Sets and Relations: Let $A=\{1,3,4,6,9\}$ and $B=\{2,4,5,8,10\}$. Let $R$ be a relation defined on $A\times B$ such that $R=\{((a_1,b_1),(a_2,b_2)):a_1\le b_2\text{ and }{b}_1\le a_2\}$. Then the number of elements in the set $R$ is

Unlocking Relations

Welcome, future engineers! Today, we are going to peel back the layers of a classic JEE Advanced problem on relations. It might look like a daunting task of counting hundreds of possibilities, but as we will see, the beauty of mathematics lies in finding the hidden simplicity within complexity.

Let us embark on this journey together.

Deconstructing the Relation

We are given two sets, $A=\{1,3,4,6,9\}$ and $B=\{2,4,5,8,10\}$. We define a relation $R$ on the Cartesian product $(A\times B)\times (A\times B)$.

The relation connects a pair $((a_1,b_1),(a_2,b_2))$ if and only if $a_1\le b_2$ and $b_1\le a_2$. At first glance, this looks like a tangled web of variables.

However, the condition $a_1\le b_2$ only involves $a_1$ and $b_2$, while $b_1\le a_2$ only involves $b_1$ and $a_2$. These two conditions are completely independent.

Because they are independent, the total number of elements in $R$ is simply the product of the number of valid pairs for each condition:

The Art of Systematic Counting

Now, let us tackle the first condition: $a_1\le b_2$. We iterate through each $a_1\in A$ and count the valid $b_2\in B$:

If $a_1=1$, then $b_2\in \{2,4,5,8,10\}$ (5 cases). If $a_1=3$, then $b_2\in \{4,5,8,10\}$ (4 cases). If $a_1=4$, then $b_2\in \{4,5,8,10\}$ (4 cases). If $a_1=6$, then $b_2\in \{8,10\}$ (2 cases). * If $a_1=9$, then $b_2\in \{10\}$ (1 case).

Summing these up, we get $5+4+4+2+1=16$. We have successfully counted 16 valid pairs for the first condition.

The Second Condition

Next, we repeat this for the second condition: $b_1\le a_2$, where $b_1\in B$ and $a_2\in A$:

If $b_1=2$, then $a_2\in \{3,4,6,9\}$ (4 cases). If $b_1=4$, then $a_2\in \{4,6,9\}$ (3 cases). If $b_1=5$, then $a_2\in \{6,9\}$ (2 cases). If $b_1=8$, then $a_2\in \{9\}$ (1 case). * If $b_1=10$, there are no elements in $A$ such that $10\le a_2$ (0 cases).

Summing these up, we get $4+3+2+1+0=10$. We have 10 valid pairs for the second condition.

We have found 16 ways to satisfy the first condition and 10 ways to satisfy the second. Since these are independent, we multiply them:

There are exactly 160 elements in the relation $R$. See how the complexity vanished once we identified the independence? Keep this mindset, and you will conquer any problem JEE throws at you!