Animated Solution for Mathematics - Sets and Relations: Let $R$ be a relation from the set $\{1,2,3,\dots ,60\}$ to itself such that $R=\{(a,b):b=pq,\text{ where }p,q\ge 3\text{ are prime numbers}\}$. Then, the number of elements in $R$ is :

A Journey into Counting

Welcome, fellow explorers of mathematics! Today, we are going to dissect a problem that at first glance seems like a simple counting exercise, but it is actually a beautiful lesson in systematic thinking and the power of the Fundamental Counting Principle.

We are looking at a relation $R$ defined on the set $S=\{1,2,3,\dots ,60\}$. The relation is defined by the condition that for any pair $(a,b)\in R$, $b$ must be the product of two prime numbers $p$ and $q$, both of which are greater than or equal to 3.

Our goal is to find the total number of elements in this relation.

The Freedom of $a$

Let's start by looking at the first component of our ordered pair, $a$. The problem statement tells us that $R$ is a relation from $S$ to $S$.

It places absolutely no restrictions on $a$ other than that it must be an element of $S$. Since $S$ contains all integers from 1 to 60, we have exactly 60 independent choices for $a$.

Whether $b$ is 9, 15, or 57, $a$ can be any of these 60 numbers. This independence is the key to our calculation.

The Hunt for $b$

Now, let's turn our attention to $b$. We are told $b=pq$, where $p,q\ge 3$ are prime numbers and $b\le 60$.

To find all distinct values of $b$, we must be systematic. We will iterate through possible values of $p$, and for each $p$, find the valid values of $q$.

To avoid the trap of double-counting, we will enforce the condition $p\le q$.

Case 1: Let $p=3$. The condition $b\le 60$ becomes $3q\le 60$, which simplifies to $q\le 20$. The primes $q$ that are $\ge 3$ and $\le 20$ are $\{3,5,7,11,13,17,19\}$.

This gives us 7 distinct values for $b$: $\{9,15,21,33,39,51,57\}$.

Case 2: Let $p=5$. We need $q\ge 5$ (to maintain $p\le q$). The condition $b\le 60$ becomes $5q\le 60$, or $q\le 12$.

The primes $q$ that are $\ge 5$ and $\le 12$ are $\{5,7,11\}$. This gives us 3 distinct values for $b$: $\{25,35,55\}$.

Case 3: Let $p=7$. We need $q\ge 7$. The condition $b\le 60$ becomes $7q\le 60$, or $q\le 8.57$.

The only prime $q$ that satisfies $7\le q\le 8.57$ is $q=7$. This gives us 1 distinct value for $b$: $\{49\}$.

If we try $p=11$, the smallest product is $11\times 11=121$, which is well beyond our limit of 60. So, we have exhausted all possibilities.

The Synthesis

We have found the number of distinct values for $b$ by summing the counts from our cases:

These 11 values are the only possible values for $b$ that satisfy the condition $b=pq$ with $p,q\ge 3$. Now, we return to the Fundamental Counting Principle.

We have 60 choices for $a$ and 11 choices for $b$. Since these choices are independent, the total number of elements in the relation $R$ is:

Isn't it elegant? By breaking a seemingly complex constraint into manageable cases, we turned a daunting problem into a simple multiplication. Keep this systematic approach in your toolkit, and you will find that even the most intimidating JEE problems start to unravel before your eyes.