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

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Roger wants to arrange three of his five books on his bookshelf. Two o  [#permalink]

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New post 13 May 2018, 11:01
5
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

42% (02:17) correct 58% (02:02) wrong based on 103 sessions

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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?

1. 12
2. 36
3. 42
4. 60
5. 128

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Re: Roger wants to arrange three of his five books on his bookshelf. Two o  [#permalink]

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New post 13 May 2018, 16:15
1
QZ 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?

1. 12
2. 36
3. 42
4. 60
5. 128


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.
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Re: Roger wants to arrange three of his five books on his bookshelf. Two o  [#permalink]

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New post 28 May 2018, 08:17
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

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Re: Roger wants to arrange three of his five books on his bookshelf. Two o  [#permalink]

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New post 14 Jul 2018, 19:19
AkshdeepS 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?

1. 12
2. 36
3. 42
4. 60
5. 128


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
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Re: Roger wants to arrange three of his five books on his bookshelf. Two o &nbs [#permalink] 14 Jul 2018, 19:19
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