Consider a $4\times 4$ grid as follows:
\begin{array}{|c|c|c|c|} \hline 1 &2&3&4\\ \hline 5&6&7&8\\ \hline 9&10&11&12\\ \hline 13&14&15&\\ \hline \end{array} Each number 1-15 represents a tile, and there is a blank space where the 16 should be. We can slide tiles horizontally or vertically to fill the blank space, for instance from this starting configuration we can slide the 15-tile to the right or the 12 tile down.
Let's permute the numbers 1-15 using an even permutation $p$. It is always possible to start with
\begin{array}{|c|c|c|c|} \hline p(1) &p(2)&p(3)&p(4)\\ \hline p(5)&p(6)&p(7)&p(8)\\ \hline p(9)&p(10)&p(11)&p(12)\\ \hline p(13)&p(14)&p(15)&\\ \hline \end{array}
and shift tiles correctly so we get the original configuration. How do I prove this? (This is in Rosen's Discrete Mathematics and its Applications, where the "proof is left as a an excercise for the reader")
Suppose you changed the order of the tiles after a number of moves. There is nothing wrong assuming the blank tile always ends up in the lower right corner. The first question that comes up is: "how many moves are possible". At first it seems that there are an infinity of possibilities since the number of paths seems unlimited. This actualliy not completely true. If you move the blank tile (BT) to a certain posititon and then let ot return to its starting point following the exact same way on the way back all the tiles retake their original position. Call those null paths. Suppose now that you move the BT to position $6$ turn clockwise around the $1-2-6-5$ pivot point and return exactly the bay back we see that we executed the permutation $(1 2 5)$. It is not difficult to see that the number of effective paths corresponds to the alternating group on $15$ elements having order $653837184000$.