= Solution
For first-bin size $k$, choose its contents in $\binom6k$ ways. Each choice arises from $k!(6-k)!$ <permutations>, obtained by ordering the two bins internally. The seven ordered size choices give
$$
\begin{array}{c|r|r|r}
\text{bin sizes}&\text{partitions}&\text{orderings per partition}&\text{product}\\\hline
\{0,6\}&1&720&720\\
\{1,5\}&6&120&720\\
\{2,4\}&15&48&720\\
\{3,3\}&20&36&720\\
\{4,2\}&15&48&720\\
\{5,1\}&6&120&720\\
\{6,0\}&1&720&720
\end{array}
$$
Here $0!=1$. Each row expresses the same <binomial coefficient> identity $\binom6k k!(6-k)!=6!$. 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.
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