Animated Solution for Mathematics - Functions: Let $x$ denote the total number of one-one functions from a set $A$ with 3 elements to a set $B$ with 5 elements and $y$ denote the total number of one-one functions from the set $A$ to the set $A\times B$. Then:

Welcome, future engineer! Today, we are diving into the elegant world of combinatorics. Often, students look at problems involving functions and sets and feel overwhelmed by the sheer number of possibilities.

But here is the secret: you don't need to count every single possibility. You just need to understand the rules of the game. We are looking at one-one functions, which are the 'exclusive clubs' of the mathematical world. In a one-one function, every element in the domain must have a unique partner in the codomain. No sharing allowed!

The First Bridge

Let's start with our first set, $A$, which has $|A|=3$ elements, and set $B$, which has $|B|=5$ elements. We want to find $x$, the number of one-one functions from $A$ to $B$.

Imagine you are the first element of $A$. You look at $B$ and see 5 available seats. You pick one. Now, the second element of $A$ looks at $B$ and sees only 4 seats left. Finally, the third element sees 3 seats.

This is the fundamental principle of counting, and it leads us directly to the permutation formula:

When we compute this, we get $5\times 4\times 3=60$. That is our $x$. It represents the total number of ways to map our 3 elements into 5 distinct targets without any collisions.

The Cartesian Twist

Now, the problem throws a curveball. We are introduced to the Cartesian product $A\times B$. Do not let the notation intimidate you! The Cartesian product is simply the set of all possible ordered pairs $(a,b)$.

If $A$ has 3 elements and $B$ has 5, then for every element in $A$, there are 5 possible partners in $B$. Thus, the total number of elements in $A\times B$ is simply:

We have just expanded our target pool from 5 to 15. This is where the magic of permutations really shows its power.

The Grand Comparison

We now need to find $y$, the number of one-one functions from $A$ to $A\times B$. We are mapping the same 3 elements from $A$, but now they have 15 targets to choose from.

Using our permutation logic again:

Now, we compare $y$ and $x$. Instead of just looking at the raw numbers, let's look at the ratio:

Watch how beautifully this simplifies! We can break this down as:

Alternatively, calculating the ratio directly:

Rearranging this, we get $2y=91x$. And there it is—the elegance of mathematics revealing the answer. You have successfully navigated the logic of permutations and set theory. Keep this confidence, and carry it into your next problem!