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Eleven books consisting of 5 financial management, 4 marketi

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Eleven books consisting of 5 financial management, 4 marketi [#permalink] New post 30 Jan 2013, 22:54
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Eleven books consisting of 5 financial management, 4 marketing management and 2 system management are placed on a shelf at random order. What is the probability that the books of each kind are all together.

A. 1/1155
B. 1/1255
C. 1/1355
D. 1/1455
E. 1/1555
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jan 2013, 03:10, edited 1 time in total.
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Re: Eleven books consisting of 5 financial management, 4 marketi [#permalink] New post 31 Jan 2013, 03:25
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antoxavier wrote:
Eleven books consisting of 5 financial management, 4 marketing management and 2 system management are placed on a shelf at random order. What is the probability that the books of each kind are all together.

A. 1/1155
B. 1/1255
C. 1/1355
D. 1/1455
E. 1/1555


(Probability of an event) = (Favorable) / (Total)

The total number of ways to arrange 11 books in a row is 11!.

3 groups of books: financial management (F), marketing management (M) and system management (S), can be arranged in 3! ways: FMS, FSM, MFS, MSF, SFM, and SMF (this way all the books of each kind will be together). Now, books in F itself can be arranged in 5! ways, in M in 4! ways and in S in 2! ways. Hence, the total number of ways to arrange books so that books of each kind are together is 3!*5!*4!*2!.

P = (3!*5!*4!*2!) / 11! = 1/1155.

Answer: A.

Hope it's clear.

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Re: Eleven books consisting of 5 financial management, 4 marketi [#permalink] New post 25 Aug 2013, 20:38
Since the denominator will be 11! . And there is no way to divide 11 from numerator and denominator. Use a multiple of 11 in the denominator only answer A

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Re: Eleven books consisting of 5 financial management, 4 marketi   [#permalink] 25 Aug 2013, 20:38
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