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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