A library has two books of three copies each and three other books

Since all copies of one type of book have to remain together, simplify the problem and consider all the copies of one type of book as one unit.

Using this simplified logic, we can see that we are asked to arrange five "units".

Five different "units" (books) can be arranged in 5! ways. Think of this as a permutation or as a combination problem where order matters.

Now, the copies within each unit can be arranged among themselves in only one way. Since all the copies are identical to each other, the order of arrangement within each unit does not make a difference. This logic can be extended to each of the five units.

Therefore, the total number of ways in which all five units with different numbers of copies each can be arranged is equal to 5! x 1 x 1 x 1 x 1 x 1 = 120 ways.

ANSWER: (A)

total types of books ; 3+2 ; 5
arrangement of 5 books can be done in 5! ways ; 120
IMO A

But we can arrange copies of the book as well, no?
I though the order of copies should matter as well, since for example we have copies of book A= C1, C2 and C3
C1C2C3
and
C2C1C3
are different arangements

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