Permutations of Comics, Novels, and Magazines Without Restrictions

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SUMMARY

The discussion centers on calculating the permutations of a collection of reading materials consisting of 4 comics, 2 novels, and 3 magazines, totaling 9 items. The initial calculation presented is 9! = 362880, which considers all items as distinct. However, when the order of the comic books is fixed, the correct formula is 9! / 4! = 1782, as this accounts for the fixed arrangement of the comics while allowing for the permutations of the novels and magazines.

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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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