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Roger wants to arrange three of his five books on his

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Roger wants to arrange three of his five books on his

by swerve » Tue Jan 15, 2019 9:25 am

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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

The OA is C

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by Brent@GMATPrepNow » Tue Jan 15, 2019 9:43 am
swerve wrote: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
I'm not crazy about this question.
The official answer suggests that, although 2 books are DUPLICATES, they're still considered DIFFERENT.

Here's what I mean:
Let's let D and d represent the two duplicate books.
Let, A, B and C represent the other three books.

The official answer suggests that the arrangement ABD is different from the arrangement ABd
I'm okay with that being the case, but there should be some text that states this.

I say this because my first reaction was to assume that the duplicate books are considered identical.
Given this assumption, we need only burn one of the duplicate books (this eliminates the chances of having the 2 duplicate books in the arrangement) and then arrange 3 of the remaining 4 books in a row (can be done 24 ways)

Cheers,
Brent
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by [email protected] » Tue Jan 15, 2019 10:57 am
Hi All,

We're told that Roger wants to arrange three of his five books on his bookshelf, but two of the five books are duplicates and cannot both be selected. We're asked for the number different ways that Roger can arrange his books. The wording of this question is a bit 'quirky', but the intent is that Roger can display EITHER of the two duplicate books (but not both) and that each option would represent a unique arrangement. This question can be approached in a number of different ways, including with a modified version of Permutation...

To start, let's call the books A, B, C and X, Y (the two duplicates).

Since the two duplicates CANNOT be displayed together, we can deal with the 5 books as two groups of 4 (ABCX and ABCY).

With A, B, C and X, we would have (4)(3)(2) = 24 different arrangements.
With A, B, C and Y, we would also have (4)(3)(2) = 24 arrangements, but there would be some duplicates that matched options in the first group of 24....

Since A, B and C appear in both groups, the arrangements involving JUST A, B and C would appear in both sets (those 6 arrangements are ABC, ACB, BAC, BCA, CAB and CBA), so we cannot count them twice. Thus, the total number of arrangements would be...

24 + 24 - 6 = 42

Final Answer: C

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by Scott@TargetTestPrep » Thu Jan 17, 2019 5:53 pm
swerve wrote: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
The number of ways to select 3 books from 5 books is 5C3 = (5 x 4 x 3)/3! = 10.

Out of these 10 ways, the number of ways that 2 of the 3 books will be duplicates is 2C2 x 3C1 = 1 x 3 = 3.

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

For any 3 selected books, Roger can arrange them in 3! = 6 ways. Therefore, Roger can arrange the books in 7 x 6 = 42 ways.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
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