Bubble sorting question

This can be computed using the fact that Bubblesort removes exactly one inversion per swap operation. So the number of swaps has the same distribution as the number of inversions, which has the ordinary generating function $$ f(z) = \prod_{k=0}^{n-1} \sum_{m=0}^k z^m.$$ E.g. for $n=4$ we get $$ f(z) = (1) \left( 1+z \right) \left( 1+z+{z}^{2} \right) \left( 1+z+{z}^{2}+{z}^{3} \right) = 1+3\,z+5\,{z}^{2}+6\,{z}^{3}+5\,{z}^{4}+3\,{z}^{5}+{z}^{6}.$$ The expected number of swaps is $$ \frac{1}{n!} \left. \frac{d}{dz} f(z)\right|_{z=1}$$ or $$\left. \frac{1}{n!} \prod_{k=0}^{n-1} \sum_{m=0}^k z^m \sum_{k=0}^{n-1} \frac{\sum_{m=1}^k m z^{m-1} }{\sum_{m=0}^k z^m } \right|_{z=1} $$ which is $$ \sum_{k=0}^{n-1} \frac{\sum_{m=1}^k m }{k+1} = \frac{1}{4} n (n-1).$$

Let $n$ be the size of the list, and $k \le n(n-1)/2$ be the number of swaps. The number you are looking for is the number of ways to write $$ k = x_1+x_2+\ldots+x_n $$ such that each $x_i \in [0, i)$. Each $x_i$ corresponds to the number of spaces that the $i$-th element in the list is "bubbled" to the left. For instance, for $n=4$ and $k=2$, the five possibilities are $$ \begin{eqnarray} 2&=&0+0+0+2 \\ 2&=&0+0+1+1 \\ 2&=&0+0+2+0 \\ 2&=&0+1+0+1 \\ 2&=&0+1+1+0. \\ \end{eqnarray} $$ The symmetry noted in the question corresponds to the mapping $x_i \rightarrow i-1-x_i$ and $k\rightarrow \sum_{i}(i-1) - k = n(n-1)/2-k$.