*n*x

*m*grid. As with the first shape-counting problem, we get a two-dimensional array of problems:

Again, the easiest problems to solve are the ones in the top row and the ones in the leftmost column. As this picture illustrates, the number of rectangles in a 1 x

*m*grid is T(

*m*), the

*m*th triangular number.

Similarly, the number of rectangles in an

*n*x 1 grid is T(

*n*). It turns out that we can use the solutions to these problems to solve every other problem. There is a one-to-one correspondence between rectangles in an

*n*x

*m*grid and pairs of rectangles from the

*n*x 1 and 1 x

*m*grids.

So the number of rectangles in an

*n*x

*m*grid is T(

*n*)*T(

*m*). For the 3 x 4 grid that's T(3)*T(4) = 6*10 = 60. Notice the similarity with the first shape-counting problem. To get the solution to a given problem in our array, just multiply the solution to the leftmost problem in the row by the solution to the top problem in the column.

**Common Factors**

The way I just explained the solution is not the way I originally figured it out. To arrive at the solution I looked at a particular example, found a product that gave the number of each type of rectangle, and then factored.

So working on this problem might provide students a good opportunity to apply their factoring skills.

**Summation Notation**

Another way to solve the problem is to manipulate some nasty sigma notation. We might realize that the number of rectangles in an

*n*x

*m*grid is the sum of all the answers to these questions: how many

*i*x

*j*rectangles are there in an

*n*x

*m*grid? For example, how many 5 x 8 rectangles are there in a 12 x 17 grid? The answer is (12 - 4)(17 - 7). The number of

*i*x

*j*rectangles in a

*n*x

*m*grid is (

*n*-

*i*+ 1)(

*m*-

*j*+ 1). So the total number of rectangles in an

*n*x

*m*grid is this double summation:

If you evaluate this sum for some particular values like

*n*= 3 and

*m*= 4 you can figure out the form of the solution like we did above. Alternatively, you can simplify the summation by realizing that

and then manipulating the sum like this:

Thanks for reading! If you solved the problem a different way or have any ideas about how this task could be implemented with students, I would be excited to hear from you!

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