"Round table" permutations are considered as if the table
and chairs and people were on a rotating platform, like the 
horses on a merry-go-round.  Though this is not realistic 
for table seating, it is nevertheless the accepted way to
assume that "round table" mathematics problems are to be 
interpreted.

In building a merry-go-round with a single circle of n
horses, the number of orders of n different colored horses 
can be installed on the merry-go-round is (n-1)!
[That's because it would be n! if the merry-go-round could
only remain still, i.e., could not rotate.  But each of the 
n rotations would be a different one of the n! "still" 
arrangements. So each "still" arrangement is counted n
times among the n!, so we divide the n! by n to get (n-1)!
So when n things are arranged in a line, the number of
possible arrangements is n!, but when they are in a circle,
the number of possible arrangements is (n-1)! 

So arranging in a straight line, the formula is n!,
and at a round table, it's (n-1)!     
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Since 3 people want to sit next to each other,

we can have them sit as a trio in 3! or 6 ways.

For each of the 3! choices for the trio, we have

7 single people plus 1 trio to seat round the table.

That's 8 things

Using the (n-1)! "round table" formula, the answer is

3!(8-1)! = 3!7! = 6(5040) = 30240 

Edwin