What I am getting is that there are 10 ways of choosing the 1 pair of shoes from the 10. Since the remaining shoes can't match, there are $\left( \begin{matrix} 9\\ 6\end{matrix} \right) $ ways to choose pairs from which to select the remaining shoes, $2^6$ ways to select individual shoes from the 6 pairs, and divided by$ \left( \begin{matrix} 20\\ 8\end{matrix} \right) $ ways of selecting 8 shoes from 20. This gives us
$\dfrac {\left( \begin{matrix} 9\\ 6\end{matrix} \right)\times 10 \times 2^6 } {\left( \begin{matrix} 20\\ 8\end{matrix} \right) }$
This gives a probability of $\dfrac {1792} {4199}$. But that doesn't seem quite right, I think I am missing something. Any help would be appreciated.
Best Answer
It looks fine to me. In general there are $\binom{10}k\binom{10-k}{8-2k}2^{8-2k}$ ways to choose the $8$ shoes so as to get exactly $k$ pairs, your calculation being the case $k=1$, and a quick numerical check confirms that
$$\sum_{k=0}^4\binom{10}k\binom{10-k}{8-2k}2^{8-2k}=\binom{20}8\;.$$