Combinations and permutations with constraints

Question

In how many ways can 6 people be seated around a table for 6, if 2 of the group are always:

• together? - I solved this by doing 2! x 4!
• separate? - I do not know how to do this, the answer is 72

Can you plz explain the second part?

Answer 1

You can use the total arrangement minus the number of arrangements when the two persons sitting together to get the number of the separate arrangements. $5!-2!\times 4!$.

Answer 2

There's always this tension between people when answering "round table" questions. Your solution of $2!4!$ for the first part suggests that you are in the camp that holds that the seat positions are indistinguishable. I would personally multiply your answer by $6$, as I have never experienced such a table.

However, in the spirit of your first answer, I would say a similar process should be used. First arbitrarily seat the first person from your restricted pair, then chose one of the three seats not next to that person for other one. Then permute the other four less-fussy dining companions into the remaining four seats. Which is just $\frac 32$ times your other answer - whichever camp you are in.