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

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

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Re: Roger wants to arrange three of his five books on his bookshelf. Two o [#permalink]
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u1983 wrote:
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.

I have a doubt. Since 2 of the 5 books are duplicates (one and the same book, but two copies), why are there 2 ways of selecting 1 out of those 2 (why 2C1). Will we not end up with the same selection, no matter which out of the 2 we select?
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Roger wants to arrange three of his five books on his bookshelf. Two o [#permalink]
JeffTargetTestPrep wrote:
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.

Dear Jeff:

Can you please list out the 7 ways to select 1 duplicate and 2 unique items? I only get 3 upon listing them out and using the combinations formula:

(2C1÷2!)×(3C2)=1×3=3

When selecting duplicates, do we divide by the factorial of the total number of duplicates? For example, if there are 3 duplicates and we are selecting 2, logically there would be only 1 way to select them. Mathematically, I was taught to divide by the factorial of the total number of duplicates. However, this adjustment would result in a fraction. So, what is the correct way to adjust for duplicates mathematically?

3C2÷3! Or 3C2÷3

Lastly, when selecting 3 duplicates out of a total of 3 options, would we still divide by either 3! Or 1 as the answer already would be 1. Thank you.

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

Dear Jeff:

Can you please list out the 7 ways to select 1 duplicate and 2 unique items? I only get 3 upon listing them out and using the combinations formula:

(2C1÷2!)×(3C2)=1×3=3

When selecting duplicates, do we divide by the factorial of the total number of duplicates? For example, if there are 3 duplicates and we are selecting 2, logically there would be only 1 way to select them. Mathematically, I was taught to divide by the factorial of the total number of duplicates. However, this adjustment would result in a fraction. So, what is the correct way to adjust for duplicates mathematically?

3C2÷3! Or 3C2÷3

Lastly, when selecting 3 duplicates out of a total of 3 options, would we still divide by either 3! Or 1 as the answer already would be 1. Thank you.

Posted from my mobile device

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Re: Roger wants to arrange three of his five books on his bookshelf. Two o [#permalink]
An observer would see 24 distinct arrangements.

"In how many ways can he select and arrange his books" might emphasize what's going on

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

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

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

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

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

Dear Karishma:

Is it possible to use the combinations formula in duplicate selection problems as one uses the permutations formula by dividing by the factorials of the number of duplicates? If so, what would be the correct approach for this problem? I am trying to ascertain a direct formulaic approach versus calculating the total number of permutations and subtracting the number of duplicate permutations.

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