## Arrangements of Guests Around Tables

24 guests are to be seated at three tables.
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