 Solution

For first-bin size $k$, choose its contents in $\left(\genfrac{}{}{0ex}{}{6}{k}\right)$ ways. Each choice arises from $k!(6-k)!$ permutations, obtained by ordering the two bins internally. The seven ordered size choices giveHere $0!=1$. Each row expresses the same binomial coefficient identity $\left(\genfrac{}{}{0ex}{}{6}{k}\right)k!(6-k)!=6!$. Even for $\{3,3\}$ the bins retain their first/second labels, so the number is twenty rather than ten; the braces denote bin sizes in the question's convention, not interchangeable bins.