In how many different ways can persons A,B,C,D and E be seated around two equal round tables so that no table remains empty?

Question

Persons A and E never sit at the same table.

I'm guessing you could describe this in terms of Stirling numbers but I'm basically lost from there.

My guess is that it's s(5,2) - the number of occurrences in which both A and E occur.

Answer 1

Consider case $n$: First table has $n$ occupied places, and second table has $5-n$ occupied places.

Number of ways to choose $n$ persons for first table is $\binom{5}{n}$.

Number of ways to seat $n$ persons around round table is $(n-1)!$. To prove it, we number occupied seats starting from seat occupied by first person (we can use alphabet rule to select first person). There are $n-1$ candidates for second seat, $n-2$ candidates for third seat, and so on.

Total number of ways for case $n$ is $A_n=\binom{5}{n}\cdot(n-1)!\cdot(4-n)!$. Last factor is number of ways for second table.

$$A_n=\frac{5!(n-1)!(4-n)!}{n!(5-n)!}=\frac{120}{n(5-n)}$$

Total number of ways is $$\sum_{n=1}^4 A_n=100$$