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Updated on: 31 Jan 2013, 04:10
Originally posted by antoxavier on 30 Jan 2013, 23:54.
Last edited by Bunuel on 31 Jan 2013, 04:10, edited 1 time in total.
Last edited by Bunuel on 31 Jan 2013, 04:10, edited 1 time in total.
RENAMED THE TOPIC.
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (02:33) correct
34%
(02:27)
wrong
based on 176
sessions
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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
A. 1/1155
B. 1/1255
C. 1/1355
D. 1/1455
E. 1/1555
31 Jan 2013, 04:25
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antoxavier
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
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.
P.S. P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3 (the name of a topic MUST be the first 40 characters (~the first two sentences) of the question).
General Discussion
25 Aug 2013, 21:38
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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
Posted from GMAT ToolKit
Posted from GMAT ToolKit 
