Animated Solution for Mathematics - Permutations and Combinations: A rectangle with sides of length $(2m-1)$ and $(2n-1)$ units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is

Unlocking the Grid

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are peeling back the layers of a grid to reveal the hidden symmetry of numbers.

When you look at a rectangle divided into unit squares, it is easy to get lost in the sheer number of possible shapes. But in the JEE Advanced arena, we do not count blindly. We look for the underlying structure.

The Boundary Perspective

Imagine you are standing before this grid. It has a width of $2m-1$ and a height of $2n-1$.

The first instinct is to count the squares, but that is a path to madness. Instead, let us focus on the boundaries—the lines that define these rectangles.

A rectangle of width $W$ is defined by two vertical lines. If our total width is $2m-1$, we have exactly $2m$ vertical lines. Similarly, for a height of $2n-1$, we have $2n$ horizontal lines.

Every rectangle in this grid is uniquely determined by selecting two vertical lines and two horizontal lines. This is our fundamental building block.

The Parity Trap

Now, here is where the problem gets interesting. We are not looking for all rectangles; we are looking for those with odd side lengths.

Let the vertical lines be at positions $i$ and $j$ (where $1\le i<j\le 2m$). The width of the rectangle is $|i-j|$. For this width to be odd, the difference between $i$ and $j$ must be odd.

Think about the arithmetic of parity: - Odd $-$ Odd $=$ Even - Even $-$ Even $=$ Even - Odd $-$ Even $=$ Odd

To get an odd width, we are forced into a specific choice: we must select one line with an odd index and one line with an even index. This is the 'Aha!' moment. The grid is not just a collection of lines; it is a collection of odd-indexed and even-indexed boundaries.

The Counting Principle

Let us count our options. In our set of $2m$ vertical lines, exactly half are odd-indexed ($1,3,\dots ,2m-1$) and half are even-indexed ($2,4,\dots ,2m$).

That gives us $m$ odd lines and $m$ even lines. To form an odd width, we choose one from the $m$ odd lines and one from the $m$ even lines.

By the fundamental counting principle, the number of ways to choose the vertical boundaries is $m\times m=m^2$.

We apply the exact same logic to the horizontal lines. We have $2n$ horizontal lines, with $n$ odd-indexed and $n$ even-indexed.

To ensure an odd height, we must choose one odd-indexed horizontal line and one even-indexed horizontal line. This gives us $n\times n=n^2$ ways.

Since the choice of vertical lines and horizontal lines are independent events—choosing the width does not restrict our choice of height—we simply multiply the possibilities.

The total number of rectangles with odd side lengths is the product of our vertical choices and our horizontal choices:

Look at the beauty of that result. It is simple, symmetric, and derived entirely from the parity of the grid lines.

When you encounter problems like this in the exam, do not panic. Break the geometry down into its constituent lines, identify the parity condition, and let the counting principle do the heavy lifting. You have mastered the grid. Keep this clarity, and you will conquer any problem that comes your way.