# Solution to 4x4 Question

Let's work this out the same way as in the 3x3 case, starting with calculating the number of latin squares of standard form, and then multiplying by the number of re-arrangements of the rows and columns. Let's start with the elbow cell. There are, this time, three choices for the elbow cell, and once we have filled in the elbow cell there is only one way to finish the row. This is because at least one of the numbers 3 and 4 hasn't been assigned as the elbow cell and can't be under itself, so has to be in the only other remaining cell, and when we are done with that there is only one remaining cell and remaining number.

This is where things get tricky. If we chose a 3 or a 4 for the elbow cell, the rest of the choices have been made for us, because there is only one way to fill in

``` 1234 2341 3 4 ```
or
``` 1234 2413 3 4 ```
on the other hand, there are two ways to fill in
``` 1234 2143 3 4 ```
there is
``` 1234 2143 3412 4321 ```
and
``` 1234 2143 3421 4312 ```
so we get a total of four latin squares of standard form.

Now we are, again, left to count the re-arrangements of these latin squares of standard form. There are 4! ways to permute the columns (and thus re-order the first row) and there are then 3! ways to arrange the remaining rows. Thus, there are a total of 4*4!*3!=576 4x4 latin squares.

This work was made possible through a grant from the National Science Foundation.