Roger wants to arrange three of his five books on his bookshelf. Two o

Question

Roger wants to arrange three of his five books on his bookshelf. Two of the five books are duplicates and can not both be selected. In how many different ways can Roger arrange his books?

(A) 12
(B) 36
(C) 42
(D) 60
(E) 128

push12345 · Accepted Answer

3 different 2 same books

1) when all different selected
=3!=6

2) when one of same and 2 from different selected
=2C1*3C2*3!= 36

Total=42

Posted from my mobile device

u1983 · Answer

Case 01 : when none of the duplicates are selected.
3 books remained(As 2 of the 5 are duplicates).
Hence, the number of ways 3 books can be selected from 3 is 3C3=1

Case 02: when one of the duplicates are selected. 
Total number of ways 
=The number of selections of the books that are completely diff from each other and the duplicate ones * The number of selections of the duplicate books
=The number of ways 2 books can be selected from 3 * The number of ways 1 books can be selected from 2 
=3C2 * 2C1 
= 3*2
= 6
Hence, total number of selection of 3 books = Case 01 selections + Case 02 selections = 6+1=7

Now 3 books can be arranged on 3P3 ways among themselves. i.e., 3! ways= 6 ways

Hence , the total number of arrangements = 7*6 = 42 ..... Thus I would go for  option C .

JeffTargetTestPrep · Answer

The number of ways to select 3 books when the two duplicates are selected is 2C2 x 3C1 = 1 x 3 = 3.

The number of ways to select 3 of 5 books is 5C3 = (5 x 4 x 3)/3! = 10.

So there are 10 - 3 = 7 ways to select 3 books when the two duplicates are not selected.

Since we can organize the 3 books in 3! = 6 ways,so we can select and organize the 5 books in 7 x 6 = 42 ways.

Answer: C

WheatyPie · Answer

It's not clear to me that swapping between the duplicates actually produces distinct arrangements. I disagree with the answer as the question is phrased.

AnirudhaS · Answer

check this answer out, https://gmatclub.com/forum/roger-wants- ... l#p2094506 If you still are not convinced, let me know which step and I will try to help.

KarishmaB · Answer

I am not sure whether "duplicate" would imply "identical". I would think it would and hence I would assume that there are 4 books and we have to pick 3 from them and arrange them to get 4C3 * 3! = 24 arrangements. 
But that is not there in the options.

So then I would consider that the duplicate books may not be identical (say one is marked or written on etc). In that case, there are 5 picks out of 3 can be picked in 5C3 = 10 ways. But the ways in which both duplicates are picked are not allowed. We can pick both duplicates in 3C1 ways (pick both duplicates and 1 more our of remaining 3 books). This gives us 3 ways. So of the 10 selections, 3 are not allowed which means 7 are allowed. 
The 3 selected books can be arranged in 3! = 6 ways.

Hence total arrangements = 7 * 6 = 42

Answer (C)

bumpbot · Answer

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Roger wants to arrange three of his five books on his bookshelf. Two o

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