Distinguish circular rotations from numbered seats

Group each family as F₁ or F₂ and label the single men M₁, M₂, M₃. Arrange these five distinguishable objects in a circle in $(5-1)!$ ways, using one man as a reference. Each family has $3!$ internal orders. There are five circular gaps, including the gap across the chosen reference; choose two different gaps for the women and arrange them in $2!$ ways. Thus $(5-1)!\,3!\,3!\binom52\,2!=17280$.

The source marks this “Not recommended”. Without separating the women, seven circular objects give $(7-1)!\,3!\,3!=25920$. If the two women form a block, six circular objects give $(6-1)!\,3!\,3!\,2!=8640$. Hence $25920-8640=17280$. As the source warns, subtracting a single together-case establishes separation here because only two objects are being separated; it does not generalise directly to pairwise separation of three or more.

Each family is a consecutive parent–child–parent block, with $2!=2$ possible parent orders. There are seven objects around a circle, so $(7-1)!\times2\times2$ circular arrangements. There are 11 individual people and 11 numbered seats. After using M₁ as the circular reference, assign its seat in 11 ways: $(7-1)!\times2\times2\times11=31680$. Part (ii) does not retain part (i)’s condition separating the two single women.