MHB Permutations of Comics, Novels, and Magazines Without Restrictions

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The discussion revolves around calculating permutations of a mixed pile of 4 comics, 2 novels, and 3 magazines. The initial calculation suggests that without restrictions, the total permutations are 9! or 362,880. However, when considering the fixed order of the comic books, the correct approach is to divide by 4! to account for their unchanging arrangement. This leads to a revised total of 1,782 permutations. The question's wording is noted as confusing, contributing to the uncertainty in finding the correct answer.
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There are three separate bundles of reading material comprising of 4 comics, 2 novels and 3 magazines. They are placed together to form one pile.

In how many ways can this be done if there are no restrictions on where the individual items are to be placed?

I say 9! = 362880

Determine the number of permutations if the order of the comic books in each bundle does not change

I'm not sure what they mean by 'each bundle' here
 
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This question is worded really strangely.

If you're arranging these into one pile and the order of the comic books doesn't change then I would say it's [math]\frac{9!}{4!}[/math]. Dividing by $4!$ accounts for all the different ways you can order the comic books but keep everything else the same. Since we want just one order I would say this is the answer but again this question is worded in a way where I'm not sure.
 
Jameson said:
This question is worded really strangely.

I agree, the answer given is 1782, I've been trialing different answers to arrive there with no luck.
 

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