$7$ people sitting at a round table with $2$ next to each other My question is:

$7$ people are sitting at a round table. $2$ of them must always be sitting next to each other. How many possible ways are there?

I thought it would be $(7-1)! \cdot 2!$ for the $2$ people but apparently the answer is $(6! \cdot 2!)/6 = 5! \cdot 2$.
Can someone explain why we divide by $6$ as well?
 A: Looking at it another way, arrange the $5$ people that don't care where they sit, but leave two spaces open somewhere at the table next to each other. Then decide whether Buddy A sits to the left of Buddy B, or his right.
A: Suppose the two people who sit next to each other are Alexandra and Benjamin.  Seat Alexandra.  It does not matter where we seat her since only the relative order of the people matters in a circular arrangement.  There are two ways we could seat Benjamin, to Alexandra's left or her right.  The remaining five people can be seated in the remaining five seats in $5!$ ways as we proceed clockwise around the table from Alexandra. Hence, there are indeed $2!5!$ admissible seating arrangements.
To address your question:  Treat Alexandra and Benjamin as a block.  Then we have six objects to arrange around the round table, the block and the other five people.  In a linear arrangement, the six objects could be arranged in $6!$ ways.  Since only the relative order matters in a circular arrangment, we divide by $6$ since there are six possible starting points for a linear arrangement corresponding to the same circular arrangement.  Hence, there are
$$\frac{6!}{6} = 5!$$
ways to arrange six distinct objects in a circle up to rotation.  Since the objects within the block can be arranged in $2!$ ways, there are
$$\frac{6!}{6} \cdot 2! = 5!2!$$
arrangements of the seven people at a round table if two particular people must be adjacent.
