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A library has six identical copies of a certain book. At any given time, some of these copies are at the library and some are checked out. How many different ways are there for some of the books to be in the library and the rest to be checked out if at least two books are in the library and at least two books are checked out? (The books should be considered indistinguishable.)

 Dec 13, 2021
 #1
avatar+203 
+1

1 book is in the library, 1 book is checked out.

 

That leaves 4 books that can either be checked out or in the library.

 

Thus, there are 24 = 16 ways for the books to by arranged.

 

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 Dec 13, 2021
edited by AlgebraGuru  Dec 13, 2021
 #2
avatar+2363 
+1

AG, your answer is absurd!

You need to learn to read the question and understand what it is asking. Blindly throwing random equations at the problem rarely results in a correct solution.

 

The answer to this question is three (3) arrangements.

The books are identical, so the only distinguishable difference is the number of books.

 

Of the six books, at least two (2) books are checked-out and at least two (2) books are in the library. So the library can have a set of two, or a set of three, or a set of four books. 

 

The quantity is the only distinguishable arrangement characteristic of the sets. 

 

 

GA

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GingerAle  Dec 15, 2021

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