anonymous
  • anonymous
stuck on a permutations question - A clerk at a bookstore is restocking a shelf of best-selling novels. He has 5 copies each of 3 different novels. How many different ways can he arrange the books on the shelf?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
figured it out , 15! / 5 ! 5! 5! with cancellation
anonymous
  • anonymous
It depends on how many books fit the shelf. If he can fit all of them on the shelf, then this is the same as the famous mississippi problem. only instead of letters you have a type of book. If he can only fit say, 3 books on a shelf, then the problem is somewhat more complicated.
anonymous
  • anonymous
\[\checkmark\]

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anonymous
  • anonymous
i think 6 he can arrange them 6 different ways i think
anonymous
  • anonymous
Imagine of all 15 books were unique. you would then have 15! different combinations. for each combination, though there are 3 sets of 5! equivalencies when you take into account 3 books and 5 copies. so 15!/(5!5!5!)
kropot72
  • kropot72
If n given things can be divided into c classes of alike things differing from class to class, then the number of permutations of these things taken all at a time is: \[\frac{n!}{n _{1}!n _{2}!....n _{c}!}\] where \[n _{1}+n _{2}+.......n _{c}=n\] So @Litovel has the correct answer.

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