smc

Suppose there are more men than women; then there are between zero and two women. If there are no women, the pair is $(0,5)$. If there is one woman, the pair is $(2,5)$. If there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $(4,5)$ and $(3,4)$. All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so $\boxed{8}$.