There are to be 10 guests around the first table, 8 guests around the third table and 6 guests around the last table.

We are concerned to ask two questions.

In how many ways can the guests be partitioned into groups with a particular group sitting around a particular table?

How many seating arrangements are there?

To answer the first question, we can choose the 10 people to sit around the first table in

\[\frac{24}{10! 14!}\]

ways.There are 14 people remaining. We can choose the 8 people to sit around the second table in

\[\frac{14}{8! 6!}\]

ways.There are 6 people remaining. We can choose the 6 people to sit around the third table in

\[\frac{6}{6! 0!}=1\]

way.We can separate the guests into groups in

\[\frac{24!}{10! 14!} \times \frac{14!}{8! 6!} \times \frac{6!}{6! 0!}\]

As for the number of seating arrangements, the ten guests around the first table can be selected and seated in \[8!\]

ways, so there are \[\frac{24}{10! 14!} \times 10! = \frac{24!}{14!}\]

ways to select and seat them.Then the eight guests around the second table can be seated in

\[8!\]

ways, so there are \[\frac{14}{8! 6!} \times 8! = \frac{14!}{6!}\]

ways to select and seat them.Finally the six guests around the second table can be seated in

\[6!\]

ways, so there are \[\frac{6!}{6!! 0!} \times 6! = 6!\]

ways to select and seat them.Multiplying these together:

\[\frac{24!}{14!} \times \frac{14!}{6!} \times 6! = 24! \]

ways to select and seat them