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shelrod007
Joined: 23 Jan 2013
Last visit: 16 Sep 2016
Posts: 99
Given Kudos: 41
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE:Information Technology (Computer Software)
Updated on: 15 Aug 2024, 02:19
Originally posted by shelrod007 on 07 Feb 2015, 19:46.
Last edited by Bunuel on 15 Aug 2024, 02:19, edited 2 times in total.
Last edited by Bunuel on 15 Aug 2024, 02:19, edited 2 times in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
51% (02:10) correct
49%
(02:30)
wrong
based on 412
sessions
History
Date
Time
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Not Attempted Yet
Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?
(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!
(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!
22 Feb 2015, 13:31
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icetray
Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?
(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!
66 presentations can be ordered in 66! ways. We want only those sequences which have Ruth - Sam - Matt ordering in them.
Ruth's, Sam's and Matt's presentations can be ordered in 3! = 6 ways in those 66! cases: RSM, RMS, MRS, MSR, SRM and SMR. So, 1/6th of 66! would have RSM sequence in them, 1/6th of 66! would have RMS sequence in them, and so on. We need only RSM sequence, which as we deduced would appear on 1/6th of 66!, so the answer is 66!/6.
Answer: E.
General Discussion
07 Feb 2015, 20:24
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shelrod007
hi shelrod..
lets take a smaller value , 4 A, B, C and D.. there are 4 people in which there is a requirement of A speaking before B and rest can come anywhere..
so it can be ..A..B.. but not ..B..A..
now if we fix two others say at 1st and 2nd.. we will have C,D,A,B and C,D,B,A, out of these only one is correct.. similarily all other scenarios will have one case where A will be befor B and one wher B before A..so only one will be correct..
in this Q, there are three people where restrictions have been put.. R>S>M...rest can be anywhere..
if u take a fixed position for remaining 63, these three people can be placed in 3! ways amongst each other. out of these 3!, only one is correct where r>s>m..
similarily for all possible 66! ways, in which 63 people can be fixed and these 3 people can be fixed only in 1 way out of 3! possible ways..
this is the reason we divide 66! by 3! to cater for the restriction put..
