Animated Solution for Mathematics - Functions: The range of the function $f(x)={}^{7-x}{P}_{x-3}$ is

Unlocking the Range

Welcome, fellow traveler on the JEE journey! Today, we are going to dissect a problem that might look like a simple algebraic expression, but it is actually a gateway into the elegant world of combinatorics.

We are tasked with finding the range of the function $f(x)={}^{7-x}{P}_{x-3}$.

The Rules of the Game

Before we calculate anything, we must understand the constraints. A function is only as strong as its domain. For the permutation ${}^n{P}_r$ to exist, we are bound by three ironclad laws:

1. The Capacity Constraint: You cannot select more items than you have. Thus, $n\ge r$.

2. The Non-Negativity Constraint: You cannot select a negative number of items. Thus, $r\ge 0$.

3. The Integrality Constraint: In the context of basic permutations, $n$ and $r$ must be whole numbers ($n,r\in \mathbb{W}$).

Let us apply these to our function where $n=7-x$ and $r=x-3$.

Defining the Domain

First, let us tackle the capacity constraint: $n\ge r$. Substituting our values, we get:

Adding $x$ to both sides and adding $3$ to both sides, we find $10\ge 2x$, which simplifies beautifully to $x\le 5$.

Next, the non-negativity constraint: $r\ge 0$. This means $x-3\ge 0$, or simply $x\ge 3$.

Combining these, we see that $3\le x\le 5$. Since $x$ must be an integer to keep $n$ and $r$ as whole numbers, our domain is restricted to the set $x\in \{3,4,5\}$.

This is the only playground where our function exists!

The Mapping

Now, the thrill of discovery! We simply map our domain to the range by evaluating $f(x)$ for each valid $x$.

For $x=3$: Arranging zero items from four is a classic case—there is only $1$ way to do it. So, $f(3)=1$.

For $x=4$: Arranging one item from three is straightforward; there are $3$ ways. So, $f(4)=3$.

For $x=5$: Arranging two items from two is $2!$, which is $2\times 1=2$. So, $f(5)=2$.

We have successfully mapped our domain $\{3,4,5\}$ to the outputs $\{1,3,2\}$.

The range of our function is the set of these outputs, which, when ordered, gives us $\{1,2,3\}$.

Isn't it satisfying? We started with a daunting permutation expression and, by respecting the constraints of the math, we distilled it into a simple, elegant set. Keep this mindset—always look for the constraints first, and the solution will follow.