Animated Solution for Mathematics - Sets and Relations: Let $A=\{1,2,3,4\}$ and $R$ be a relation on the set $A\times A$ defined by $R=\{((a,b),(c,d)):2a+3b=4c+5d\}$. Then the number of elements in $R$ is

Welcome, future engineer! Today, we are going to peel back the layers of a seemingly simple problem. At first glance, it looks like a dry exercise in set theory, but I want you to see it for what it truly is: a delicate balancing act.

We are given a set $A=\{1,2,3,4\}$ and a relation $R$ defined on $A\times A$ by the condition $2a+3b=4c+5d$. Our mission is to find the number of elements in $R$.

This means we are looking for the number of quadruplets $(a,b,c,d)$ that satisfy this equation, where every single variable is trapped within the confines of our set $A$.

Before we start throwing numbers at the equation, we must understand the "playing field." We have the expression $2a+3b$ on the left and $4c+5d$ on the right.

Consider the expression $2a+3b$. Since $a$ and $b$ are at least $1$, the minimum value is $2(1)+3(1)=5$. Since $a$ and $b$ are at most $4$, the maximum value is $2(4)+3(4)=20$.

This is a massive revelation! It tells us that the expression $4c+5d$ must also fall within the range $[5,20]$. If $4c+5d$ is less than $5$ or greater than $20$, it simply cannot be part of our relation.

Now, let's identify our targets. We need to find all pairs $(c,d)$ such that $5\le 4c+5d\le 20$. Let's test them systematically:

- If $c=1,d=1$, then $4(1)+5(1)=9$. - If $c=2,d=1$, then $4(2)+5(1)=13$. - If $c=1,d=2$, then $4(1)+5(2)=14$. - If $c=3,d=1$, then $4(3)+5(1)=17$. - If $c=2,d=2$, then $4(2)+5(2)=18$. - If $c=1,d=3$, then $4(1)+5(3)=19$.

Any other combination, like $c=4,d=1$, gives $16+5=21$, which is outside our range. So, our target sums are $9,13,14,17,18,$ and $19$.

Now, we play the game of matching. For each target sum $S$, we solve $2a+3b=S$ for $a,b\in \{1,2,3,4\}$.

Target $S=9$: We need $2a+3b=9$. If $b=1$, $2a=6$, so $a=3$. This is valid! We have one pair: $(3,1)$.

Target $S=13$: We need $2a+3b=13$. If $b=3$, $2a=4$, $a=2$. This is valid! We have one pair: $(2,3)$.

Target $S=14$: We need $2a+3b=14$. If $b=2$, $2a=8$, $a=4$. If $b=4$, $2a=2$, $a=1$. We have two pairs: $(4,2)$ and $(1,4)$.

Target $S=17$: We need $2a+3b=17$. If $b=3$, $2a=8$, $a=4$. We have one pair: $(4,3)$.

Target $S=18$: We need $2a+3b=18$. If $b=4$, $2a=6$, $a=3$. We have one pair: $(3,4)$.

Target $S=19$: We need $2a+3b=19$. Testing values shows no integer solution for $a\in \{1,2,3,4\}$.

By breaking the problem down into these manageable cases, we have systematically uncovered every single valid element of the relation $R$. We sum the valid pairs found for each target:

This problem teaches us a vital lesson for JEE Advanced: never rush into calculations. Always pause, define your boundaries, and create a strategy.

The final answer is 6. Keep this clarity of thought, and you will conquer any problem that comes your way!