\item Given the set of list $x_1, x_1, x_2, x_2, \cdots, x_k, x_k$, we know that there are $2k$ elements in the list However, there are still only $k$ unique elements. Therefore, as you arrange these symbols in different orders, it will look no different if the first two $x_1$ (or any other equal terms) to be rearranged, though that may be considered an option. That is where the equation for $\frac{2k!}{2^k}$ fits. Given that there are $2k$ terms in the list, there would likely be $2k!$ possibilities for them to be rearranged. However, this is divided by $2^k$ to account for the fact that switching two equivalent terms will not appear to produce a different order. The logic behind this is that for every term from $x_1$ to $x_k$ there will need to be a division by 2 for each of the possible rearrangements.