How many nonrepeating drawings can be made up by 5 squares?

Question

Nikki moves five pictures around on a table and then steps back to look at the arrangement.  He is arranging the photo graphs for a page in the scho yearbook. The five pictures that he is working with are all the same size square. If each photo has to share at least one side with another photo, what are all the different ways he could place the pictures on the page?

Answer 1

Hi Claudia:

Here's how to tackle this kind of "counting" problem.

First, how many possibilities do we have? Well, we have 5 square photos, each with 4 sides, for a total of 20 sides.

Second, are there any restrictions on choosing a side? Yes, any side can go with any other side but only on a different photo, that is, any side on a photo can't pair up with any of the other 3 sides on the same photo.

So, we have a 2 step choice process: 1. choose the next photo and 2. choose a side from the photo to be next to a side from another photo.

The question doesn't mention the orientation of the photos (sideways, upside down, etc.), so we'll ignore that, meaning a photo can go in any orientation.

For the first photo, we have 5 choices

For the 2nd photo, we have 4 choices (since 1 has been taken.)

Since these are independent of each other, there is a total of 5 x 4 = 20 choices

For the 3rd photo, we have 3 choices, for a total of 5 x 4 x 3 = 60 choices

. . . . . 4th . . . . . .. . . . . . . 2 . . . . . . . . . . . . . . . . 5 x 4 x 3 x 2 = 120 choices

and . . . 5th . . . . . . . . . . . . . 1 choice . . . . . . . . . . . .5 x 4 x 3 x 2 x 1 = 5! = 120 choices

Now, let's see about the sides.

For the first photo, there is only 1 choice: Put the photo down somewhere

. . . . second . . . . . . . . .are 4 sides on the first photo to share sides with

. . . . . third . . . . . . . . . . . . ..6 sides, 3 on each of the previous photos, to share sides with

. . . . . fourth . . . . . . . . . . . . 8 sides since with each new photo we add 4 sides but take away 2 (the ones next to each other) for a net add of 2 sides

For the fifth photo, there are 8 + 2 = 10 sides to choose from

These are also all independent events so again we multiply: 1 x 4 x 6 x 8 x 10 = 1,920

Therefore, total number of choices is 120 x 1,920 =230,400 ( a pretty big number!)