Animated Solution for Mathematics - Permutations and Combinations: The total number of functions, $f:\{1,2,3,4\}\to \{1,2,3,4,5,6\}$ such that $f(1)+f(2)=f(3)$, is equal to:

Unlocking the Constraint

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a counting problem; we are dissecting the very anatomy of a function.

When you see a problem like $f:\{1,2,3,4\}\to \{1,2,3,4,5,6\}$ with the condition $f(1)+f(2)=f(3)$, do not rush to the numbers. Pause and visualize the domain and the codomain.

You have four inputs on the left and six potential destinations on the right. Usually, in a function, every element is free to choose its destination. But here, the condition $f(1)+f(2)=f(3)$ acts like a binding contract that links three of our four domain elements into a single, interdependent trio.

The Dependency Chain

Let us look at the trio: $f(1)$, $f(2)$, and $f(3)$. They are not independent.

If you pick a value for $f(1)$ and a value for $f(2)$, the value of $f(3)$ is instantly decided. This is the core of our problem.

We cannot simply calculate the total functions ($6^4$) and subtract the invalid ones; that would be a nightmare. Instead, we must embrace the constraint and analyze the possible values of $f(3)$ to see how many ways we can satisfy the equation $f(1)+f(2)=f(3)$ for each case.

The Case Analysis

We must ask ourselves: what are the boundaries for $f(3)$? Since the codomain is $\{1,2,3,4,5,6\}$, the smallest value any function can take is $1$.

Therefore, $f(1)\ge 1$ and $f(2)\ge 1$. This implies that:

So, $f(3)$ can be $2,3,4,5,$ or $6$. Let us break this down systematically:

1. Case $f(3)=2$: We need $f(1)+f(2)=2$. The only positive integer solution is $(1,1)$. That is 1 way.

2. Case $f(3)=3$: We need $f(1)+f(2)=3$. The pairs are $(1,2)$ and $(2,1)$. That is 2 ways.

3. Case $f(3)=4$: We need $f(1)+f(2)=4$. The pairs are $(1,3),(3,1),$ and $(2,2)$. That is 3 ways.

4. Case $f(3)=5$: We need $f(1)+f(2)=5$. The pairs are $(1,4),(4,1),(2,3),$ and $(3,2)$. That is 4 ways.

5. Case $f(3)=6$: We need $f(1)+f(2)=6$. The pairs are $(1,5),(5,1),(2,4),(4,2),$ and $(3,3)$. That is 5 ways.

Do you see the elegance of the pattern? The number of ways to satisfy the condition is simply $f(3)-1$.

It is a beautiful, linear progression: $1,2,3,4,5$. Summing these up, we find the total number of ways to map the first three elements is:

The Forgotten Element

Now, I want you to take a deep breath. Many students, in their excitement to solve the constraint, stop here.

They see the number $15$ and they want to circle it. But look at the domain again: $\{1,2,3,4\}$. We have only accounted for $1,2,$ and $3$. What about $4$?

Element $4$ is the 'free agent' of this problem. It has no constraints and is not involved in the sum. It can map to any of the $6$ elements in the codomain.

This is where the Fundamental Principle of Counting comes to our rescue. We have $15$ ways to handle the trio $(f(1),f(2),f(3))$ and $6$ independent ways to handle $f(4)$.

To find the total number of valid functions, we multiply the possibilities:

There it is. The complexity collapses into a simple, elegant result. We navigated the constraints, identified the dependency, mapped the cases, and remembered the forgotten element.

This is the essence of JEE problem-solving: staying calm, being systematic, and never losing sight of the entire domain. You have mastered this logic. Now, go forth and apply this same systematic rigor to every problem you face.