How many ways can it be arranged on a shelf? : GMAT Problem Solving (PS)
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# How many ways can it be arranged on a shelf?

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How many ways can it be arranged on a shelf? [#permalink]

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12 Jul 2011, 01:22
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Difficulty:

45% (medium)

Question Stats:

52% (01:44) correct 48% (00:43) wrong based on 31 sessions

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There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!
B) 20!/5(4!)
C) 20!/(4!)^5
D) 20!
E) 5!
[Reveal] Spoiler: OA
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Re: How many ways can it be arranged on a shelf? [#permalink]

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12 Jul 2011, 01:33
20!/((4!)^5)

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Re: How many ways can it be arranged on a shelf? [#permalink]

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12 Jul 2011, 01:35
Alchemist1320 wrote:
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!
B) 20!/5(4!)
C) 20!/(4!)^5
D) 20!
E) 5!

formula : The number of ways in which MN different items can be divided equally into M groups, each containing N objects and the order of the groups is important is = (mn)!/(n!)^m

20!/(4!)^5= C
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Re: How many ways can it be arranged on a shelf? [#permalink]

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30 Sep 2011, 03:47
Could someone explain me the rationale behind this formula ?

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Re: How many ways can it be arranged on a shelf? [#permalink]

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30 Sep 2011, 08:31
Loki2612 wrote:
Could someone explain me the rationale behind this formula ?

Posted from GMAT ToolKit

We have 5 books A,B,C,D,E and 4 copies of each. Therefore we have A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4,D1,D2,D3,D4,E1,E2,E3,E4 = 20 BOOKS

The way to rearrange 20 items is by 20x19x18x17x16....=20!

Lets keep in mind that we rearrange our 4 copies of each book by 4x3x2x1=4!

Therefore we have 5 items repeated 4 times and we need to account for the copies resulting in 20! divided by 4!x4!x4!x4!x4!

Result is option C = 20!/(4!)^5
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Re: How many ways can it be arranged on a shelf? [#permalink]

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30 Sep 2011, 11:31
Look it as number of ways of arranging AAAA BBBB CCCC DDDD EEEE books where A,B,C,D,E repeat 4 times.
hence 20!/ 4!^5
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Re: How many ways can it be arranged on a shelf? [#permalink]

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30 Sep 2011, 17:53
20! / ((4!)^5)

this division is done to avoid repetitions.

Lets say we have to figure out number of arrangements for A,B1,B2. (where B1=B2) . total arrangements for 3 letters is 3!.

A-B1-B2
A-B2-B1 - duplicate as B1=B2
B1-A-B2
B2-A-B1 - duplicate as B1=B2
B1-B2-A
B2-B1-A - duplicate as B1=B2

so to avoid duplicates we need to divide the total arrangements/ (number of similar items)! = 3!/2!

Loki2612 wrote:
Could someone explain me the rationale behind this formula ?

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Re: How many ways can it be arranged on a shelf? [#permalink]

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01 Oct 2011, 03:27
sudhir18n wrote:
Alchemist1320 wrote:
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!
B) 20!/5(4!)
C) 20!/(4!)^5
D) 20!
E) 5!

formula : The number of ways in which MN different items can be divided equally into M groups, each containing N objects and the order of the groups is important is = (mn)!/(n!)^m

20!/(4!)^5= C

thnx for the formula. please tell me what is the formula if order is not important
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Re: How many ways can it be arranged on a shelf? [#permalink]

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01 Oct 2011, 03:31
C

It's a basic formula
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Re: How many ways can it be arranged on a shelf?   [#permalink] 01 Oct 2011, 03:31
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