SOLUTION: a party twelve is to dine at three tables at a hotel. in how many ways may they split up if each table holds four

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Question 756090: a party twelve is to dine at three tables at a hotel. in how many ways may they split up if each table holds four
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that by "how many ways may they split up", that
it doesn't matter which of the three tables a group of 4 sit at,
nor does it matter which of the 4 seats they take at the tables.  
It would be a much larger answer if either or both of those 
things mattered.
 
We can choose four people from the 12 in C(12,4) ways.
For each of those ways, we can choose four people from the remaining 8 
in C(8,4) ways.  That's C(12,4)×C(8,4)
Then four from the remaining four can only be chosen in C(4,4) or 1 way.

Answer - C(12,4)×C(8,4)×C(4,4) = 495×70×1 = 34,650.

Edwin