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29 Sep 2019, 10:07
Kudos
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A
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Difficulty:
25%
(medium)
Question Stats:
77% (01:44) correct
23%
(02:15)
wrong
based on 163
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A library has two books of three copies each and three other books of two copies each. What is the number of ways in which all the books can be arranged on a shelf such that copies of the same book remain together?
(A) 120
(B) 140
(C) 160
(D) 180
(E) 320
(A) 120
(B) 140
(C) 160
(D) 180
(E) 320
Number of ways of arranging two books of three copies each and three other books of two copies each such that copies of the same book remain together will be same as that of arranging 5 distinct books.
Number of ways= 5!=120
Number of ways= 5!=120
Hovkial
30 Sep 2019, 10:44
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Hovkial
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)
