permutation-8 people forming a circle with two to be seated with each other

Question

I have one last question regarding permutation. I understand the problem and the rule of product but this problem seems to be in a different format compared with the two questions I asked before.

A committee of eight is to form a round table discussion group. In how many ways may they be seated if the 2 members are to be seated with each other?

If I understand it correctly, two persons are to choose from 6 spaces to sit,this would be 8*7=56 am I correct?

Answer 1

Form a string of the two people to be seated together which will be treated as a single person hereafter. First of all the people in this string can be permuted in $2!$ ways.

Now using circular permutations we have to place $7$ people on a circular table. Hence the number of ways for this is $6!$

Hence the total ways is $6!2!$

Update:

Let those 8 people be $A, B, C, D, E, F, G, H$. Let the special pair be of $A$ and $B$. Now consider the following two permutations do these permutations tend to differ from each other.

I guess you could have understood why I multiplied by $2!$

Answer 2

First seat a given one of the pair that have to sit together. As the table is round, all seats are equivalent, so there is only one way to do this. You have two choices for the second of the pair, then can seat the rest in $6!$ ways.