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 one book is in the library and at least one book is checked out (the books should be considered indistinguishable). Please please please help!!

Question

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 one book is in the library and at least one book is checked out (the books should be considered indistinguishable).  Please please please help!!

anonymous · Answer

Really stuck on this would really appreciate any help!

b87lar · Answer

If I cared about which book is out and which in, the number of ways the given scenario could happen is $$n=\left(\begin{matrix}6 \ 1\end{matrix}\right)+ \left(\begin{matrix}6 \ 2\end{matrix}\right)+\left(\begin{matrix}6 \ 3\end{matrix}\right)+\left(\begin{matrix}6 \ 4\end{matrix}\right)+\left(\begin{matrix}6 \ 5\end{matrix}\right) =  2^6-2=62$$Now, they say the books are indistinguishable so I think all the combinations become just one possibility each, hence there would only be 5 different ways, provided I have no way to tell the books apart:$$n^*=5$$

perl · Answer

you have 2x2x2x2x2x2 possibitilies minus something. 
because how many ways can you count these ordered pairs
< in , out, in , out, in , out,>

perl · Answer

now you need to have at least one  'in' and one 'out' . 
in = in library 
out = checked out. 
lets use 1 and 0 instead

perl · Answer

1= in library 
0 = checked out

perl · Answer

so i think you can remove one 'in' and one 'out'. then count the remaining possibilities

perl · Answer

1x1x2x2x2x2 = 16

b87lar · Answer

If the books are distinguishable then the number is 62 (see my post above), not 16.

perl · Answer

the directions say the books are not distinguishable

perl · Answer

sorry i dont understand your post b8latr

perl · Answer

maybe you could explain it

b87lar · Answer

@perl First, your calculation 1x1x2x2x2x2 = 16 would mean "the first book is always in,  the second book is always out, and the remaining 4 books can be in or out" but that's not really the setup. My point is that the if the books *were* distinguishable (yes i mean that), then the number would have been 62, which is: 2^6 - 2 (the 2 is due to the "all books in" and "all books out"). My second point answers the main question: the books are to be considered *indistinguishable* which basically tells us forget the combination business: there can only be 5 ways: 1, 2, 3, 4, 5 books out, that's it no futher distinction can be made. But maybe I misunderstand the text of the problem.

perl · Answer

ok so there are 5 books checked in or 4 books checked in or 3 or 2 or 1