Hi there! Me again. *munches*

Sorry, I still haven't finished my hamburger since the last time we count squares, I hope you don't mind that.

Anyway, we are here to see how to count rectangles of a certain size in a grid.

Consider this question: How many **3\(\times\)2 rectangles** are there in the 4\(\times\)5 grid below?

I'm a little bit picky here, note that dimensions here are expressed as **width\(\times\)height**, so in this case I only want 3\(\times\)2 rectangles (that is, a rectangle of width 3 and height 2) and not the 2\(\times\)3 ones.

If we mark an 'X' on every bottom right corner of every 3\(\times\)2 rectangle, we would see this:

Hey, that means counting the number of 3\(\times\)2 rectangles in the original grid is the same as counting how many X's are there, or in other words, it is the same as counting the number of unit squares in an 2\(\times\)4 grid.

There are \(2\times4=8\) unit squares in an 2\(\times\)4 grid, therefore, the number of 3\(\times\)2 rectangles in the original grid is 8! Hooray!

*munches*

Mmm... so in general, how many \(x\times y\) rectangles (\(x\) is the **width** of the rectangle, \(y\) is the **height** of the rectangle) are there in an \(a\times b\) grid (\(a\) is the **width** of the grid, \(b\) is the **height** of the grid), given that \(x\leqslant a\) and \(y\leqslant b\)?

Again imagine marking an 'X' on every bottom right corner of every \(x\times y\) rectangle, we will get a grid full of X's.

The width of the "X grid" is \(a-x+1\), while the height of the "X grid" is \(b-y+1\).

Since the number of \(x\times y\) rectangles in an \(a\times b\) grid equals the number of unit squares the "X grid", hence the number of \(x\times y\) rectangles in an \(a\times b\) grid is \[(a-x+1)(b-y+1)\].

*munches*

The number of \(x\times y\) (\(x\) is the

widthof the rectangle, \(y\) is theheightof the rectangle) rectangles (the exact \(x\times y\) rectangle, not including rotated variations such as \(y\times x\)) in an \(a\times b\) grid (\(a\) is thewidthof the grid, \(b\) is theheightof the grid) provided that \(x\leqslant a\), \(y\leqslant b\) is \[(a-x+1)(b-y+1)\]

This is one part of Quadrilatorics.

## Comments

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TopNewestI think another way to do it is breaking them up. For example, if we want to count the number ot rectangles in \(10×10\) grid, is first counting the number of rectangles in \(10×1\) grid then xounting the number of rectangles in \(10×2\) grid, then in \(10×3\), and so in, in this way we would be able to establish a pattern in the number of rectangles, and by arithmetic/geometric progression, we would het the answer. – Ashish Siva · 8 months, 2 weeks ago

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– Tan Kenneth · 8 months, 2 weeks ago

Yes, but in this case I only want to count rectangles of a specific size.Log in to reply

– Ashish Siva · 8 months, 2 weeks ago

We can do in that case too. Your questions are awesome.Log in to reply

problem to see my approach. – Ashish Siva · 8 months, 2 weeks ago

You may see the solution of myLog in to reply