Explain every complement and the meaning of neighbouring ride seats

The source counts subsets of distinct rides, not their visiting order or repeat rides. Thus $2^{15}-\binom{15}0-\binom{15}1=32768-1-15=32752$.

$\sum_{r=2}^{15}\binom{15}r=\binom{15}2+\binom{15}3+\cdots+\binom{15}{15}=32752$.

The nine people are distinct and the nine seat positions are distinct: $9!=362880$.

Choose three people for the first row, three from the remaining six for the second and the final three for the last. Each row has $3!$ orders. Thus $\binom93\binom63\binom33(3!)^3=362880$.

The printed count uses horizontal neighbours within a row: two pairs per row, six pairs altogether. Choose one pair in $\binom61$ ways, order Tom/mother in $2!$ ways, then arrange the remaining seven people in $7!$ ways. Hence $2!\binom61\,7!=60480$. Vertical or diagonal proximity is not counted as “next to” in this seating model.

After choosing the pair and its two orders, choose the third person in that row from 7. Choose three of the remaining 6 for one other row, leaving three for the final row, and order each other row in $3!$ ways. Thus $2!\binom61\binom71\binom63\binom33(3!)^2=60480$.

Arrange the six adults circularly in $(6-1)!$ ways. Their six gaps receive three distinct children in $6\times5\times4$ ways. Thus $(6-1)!\times6\times5\times4=14400$.

Start with $(9-1)!=40320$. All three children together contribute $(7-1)!\,3!=4320$. For exactly one adjacent pair, arrange six adults circularly in $(6-1)!$ ways; choose and order the child pair in $\binom32\,2!$ ways; choose two different adult gaps for that pair and the singleton, and assign them in $\binom62\,2!$ ways. This gives 21600. Hence $(9-1)!-(7-1)!\,3!-[(6-1)!\binom32\,2!\binom62\,2!]=40320-4320-21600=14400$.