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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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New post 12 Jul 2011, 02:22
1
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

46% (00:56) correct 54% (00:46) wrong based on 46 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!
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Re: How many ways can it be arranged on a shelf?  [#permalink]

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New post 12 Jul 2011, 02:33
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20!/((4!)^5)

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

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New post 12 Jul 2011, 02:35
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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New post 30 Sep 2011, 04:47
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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New post 30 Sep 2011, 09:31
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Loki2612 wrote:
Could someone explain me the rationale behind this formula ?

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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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New post 30 Sep 2011, 12:31
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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New post 30 Sep 2011, 18:53
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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New post 01 Oct 2011, 04:27
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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New post 01 Oct 2011, 04:31
C

It's a basic formula
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Re: How many ways can it be arranged on a shelf?  [#permalink]

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Re: How many ways can it be arranged on a shelf? &nbs [#permalink] 07 Jul 2018, 07:00
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