jmc

We pretend that the letters are all different, and we have P${}_1$A${}_1$P${}_2$A${}_2$. We have $4!$ permutations (since all the letters are different).

But how many arrangements of P${}_1$A${}_1$P${}_2$A${}_2$ (where the P's and A's are considered different) correspond to a single arrangement of PAPA (where the P's and A's are identical)? PAPA is counted in 4 different ways: as P${}_1$A${}_1$P${}_2$A${}_2$, P${}_1$A${}_2$P${}_2$A${}_1$, P${}_2$A${}_1$P${}_1$A${}_2$, and P${}_2$A${}_2$P${}_1$A${}_1$. Rather than list out the possibilities, we could have reasoned as follows: For the 2 P's, each possibility is counted $2!=2$ times, and for the 2 A's, each of these 2 possibilities is counted $2!=2$ times, for a total of $2\times 2=4$ ways. (Make sure you see why it isn't $2+2=4$ ways.) Therefore, there are 4! ways to arrange the 4 letters P${}_1$A${}_1$P${}_2$A${}_2$; this counts each arrangement of PAPA $2!\times 2!=4$ times, so we have $4!/(2!\times 2!)=\boxed{6}$ ways to arrange PAPA.