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anonymous
 3 years ago
A collection of books has 5 books on mathematics, 7 books on physics and 6 books on chemistry. In how many ways can you combine the collection on a shelf if you want the same subject to stand together?
The combinations within each subject should be, 5!, 7! and 6!, right? But how do i calculate the combinations so that they stand together?
anonymous
 3 years ago
A collection of books has 5 books on mathematics, 7 books on physics and 6 books on chemistry. In how many ways can you combine the collection on a shelf if you want the same subject to stand together? The combinations within each subject should be, 5!, 7! and 6!, right? But how do i calculate the combinations so that they stand together?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there are \(3!=6\) ways to arrange the three subjects

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so multiply all that together and then multiply by 6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Multiply which together? 5!6!7! ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Who did you choose what to divide with?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Permutations like? \[\frac{ 5!6!7! }{ (5!6!7!?) }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you are taking all the books right? don't get married to these formulas the number of ways to arrange items is 5! by the counting principle. similarly for 6 items and 7 items this is not asking you how many ways you can arrange 4 out of 10 for example

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(5!\) ways for math books, by the counting principle \(6!\) ways for the chemistry books by the counting principle and \(7!\) ways for the physics books, again by the counting principle there if you had them arranged as {math, physics, chemistry} then again by the counting principle there would be \(5!6!7!\) ways to arrange the books, but there are \(3!=6\) arrangments of the subjects, so you again need to multiply by 6 i.e. \(3!5!6!7!\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh I get it! Thank you so much! :D
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