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Litovel
Group Title
stuck on a permutations question
 A clerk at a bookstore is restocking a shelf of bestselling novels. He has 5 copies each of 3 different novels. How many different ways can he arrange the books on the shelf?
 one year ago
 one year ago
Litovel Group Title
stuck on a permutations question  A clerk at a bookstore is restocking a shelf of bestselling novels. He has 5 copies each of 3 different novels. How many different ways can he arrange the books on the shelf?
 one year ago
 one year ago

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Litovel Group TitleBest ResponseYou've already chosen the best response.0
figured it out , 15! / 5 ! 5! 5! with cancellation
 one year ago

freewilly922 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

freewilly922 Group TitleBest ResponseYou've already chosen the best response.1
\[\checkmark\]
 one year ago

KiaraMarieKlinkefus Group TitleBest ResponseYou've already chosen the best response.0
i think 6 he can arrange them 6 different ways i think
 one year ago

freewilly922 Group TitleBest ResponseYou've already chosen the best response.1
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!)
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago
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