How many ways can 10 people be seated in a row if ...
(a)there are 5 adults and 5 children, and no two adults nor two children can sit next to each other?
(b)there are 5 married couples and each couple must sit together?
2026-04-25 19:08:32.1777144112
How many ways can 10 people be seated in a row if ...
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For (a), if we denote children with a C and adults with an A, we have two possible distributions: $$C-A-C-A-C-A-C-A-C-A$$ or $$A-C-A-C-A-C-A-C-A-C$$ Now we have to consider the possibilities for adults' seats ($5!$, it is a permutation because we "use" all the elements and order is important, as I am assuming that we consider each person is different). The same would apply to children. Therefore, the number of ways in which they can sit is $2·5!·5!$.
As for (b), since couples must sit together, it is as if they were a "pack". Because of this, it is as if we just had to sit five people (instead of 10). This would also be a permutation, and the number of possible ways to sit is $5!$.
**Edit:**It has been brought to my attention in the comments that, for (b), we must also take into account the fact that every couple can sit in two different ways.