The puzzle is a popular topic for books on recreational mathematics, most of which use it as an example to illustrate the consequences of even and odd permutations. Most references to the 15-puzzle explain the impossibility of obtaining odd permutations, and many state the classical result that every even permutation is indeed possible, however complete proofs are rare. Indeed, the famous mathematicians Herstein and Kaplansky wrote in 1978 that no really easy proof seems to be known. Archer’s (1999) paper set out to rectify that deficiency. Central to this is the algebraic result that the group An with n = 15 is generated by the consecutive 3-cycles which has to be justified by suitable placements in the puzzle. We adopted Archer‘s idea, modified his enumeration and implemented a cyclic numbering of the cells where the beginning and the end of the path is the blank cell. Using this method we are able to prove the classical result rather quickly. Further, we generalize the structure of the puzzle and make clear how the result can be extended to suitable polygonal graphs.