BMW M3 Forum (E90 E92)

BMW Garage BMW Meets Register Search Today's Posts Mark Forums Read


Go Back   M3Post - BMW M3 Forum > BIMMERPOST Universal Forums > Off-Topic Discussions Board
 
Post Reply
 
Thread Tools Search this Thread
      07-30-2010, 06:36 PM   #1
Makushr1
Lieutenant
 
Makushr1's Avatar
 
Drives: 2008 335i Sedan Auto
Join Date: May 2008
Location: Texas

Posts: 480
iTrader: (2)

GMAT Quant Questions Help

It's a permutation/combination problem. I can get most of these correct, but for some reason this one is a PITA.

A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that both of them will sit next to only one other student from the group? The Answer is 10%

First, I assume you need to find out the total number of of seating arrangements. It's 5!=120.

Then, I find out how many options are available so that they sit next to only 1 other student. 5!/(3!2!)=10... Obviously either this number or the first are incorrect, since 10/120 is not 10%.

So, I tried to figure out all the possible ways that could sit, let L be Lisa, B be Bob, and x be the other students.
LBxxx
BLxxx
xBLxx
xLBxx
xxLBx
xxBLx
xxxBL
xxxLB
BxxxL
LxxxB

That is still 10. I cant quite figure out what I'm doing wrong. I think it's probably the way I'm reading the question, so hopefully some fresh eyes can help.
__________________
2008 Montego Blue E90 ll Custom K&N Dual Cone Intake ll Aluminum OEM Pedals ll 25% Tint ll M3 Spoiler

To Do (Cosmetic): Black Line lights, Lux 4.0 Angel Eyes, 19" CSL HyperBlack
To Do (Performance): Exhaust (still undecided which brand), SSTT, FMIC (unsure what type)
Makushr1 is offline  
0
Reply With Quote
      07-30-2010, 06:43 PM   #2
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

does 1 2 7 and 8 count since only one of them is sitting next to one other person from the group, or does sitting next to each other count?
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 07:08 PM   #3
Makushr1
Lieutenant
 
Makushr1's Avatar
 
Drives: 2008 335i Sedan Auto
Join Date: May 2008
Location: Texas

Posts: 480
iTrader: (2)

Quote:
Originally Posted by AW335TT View Post
does 1 2 7 and 8 count since only one of them is sitting next to one other person from the group, or does sitting next to each other count?
You may be correct. But I still need to get 10%. If those dont count, that's 6 possible positions, so I would need the denominator to be 60.

5!/2!=60, but not sure the reasoning for that...
__________________
2008 Montego Blue E90 ll Custom K&N Dual Cone Intake ll Aluminum OEM Pedals ll 25% Tint ll M3 Spoiler

To Do (Cosmetic): Black Line lights, Lux 4.0 Angel Eyes, 19" CSL HyperBlack
To Do (Performance): Exhaust (still undecided which brand), SSTT, FMIC (unsure what type)
Makushr1 is offline  
0
Reply With Quote
      07-30-2010, 07:30 PM   #4
wj4
Major General
 
Drives: E90 manual 335i
Join Date: Nov 2006
Location: Burbank, CA

Posts: 7,723
iTrader: (8)

Send a message via AIM to wj4
Just curious...where did you get the question from? I'm asking because I'm studying for the GMAT as well, taking it on August 11th. I'm using the prep book from the folks at GMAT to study and the questions come with explanations.
wj4 is offline   United_States
0
Reply With Quote
      07-30-2010, 07:49 PM   #5
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Are you sure the question doesn't specify how many seats are in a row, and that the number isn't more than five?
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 07:51 PM   #6
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

So isn't 6 10% of 60? 120 possibilities divided by 2 ppl. Why did you do 5!/(3!2!)?
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 07:52 PM   #7
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

Quote:
Originally Posted by radix View Post
Are you sure the question doesn't specify how many seats are in a row, and that the number isn't more than five?
why would that number matter it says they are all sitting next to each other.
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 07:59 PM   #8
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Quote:
Originally Posted by AW335TT View Post
why would that number matter it says they are all sitting next to each other.
Because if there are only five seats in a row there are only 4 permutations in which both of them don't sit next to one other person. I've used 'o' to mark the other person.

bloxx
lboxx
xxolb
xxobl

In each of these cases only one sits next to another person, but not both. In every other conceivable case, both of them individually must sit next to at least one another person.

loxob
boxol
oblox
olbox
xolbo
xoblo
obolo
olobo


etc...

Last edited by radix; 07-30-2010 at 08:29 PM.
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 08:11 PM   #9
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

Quote:
Originally Posted by radix View Post
Because if there are only five seats in a row there are only 4 permutations in which both of them don't sit next to one other person. I've used 'o' to mark the other person.

bloxx
lboxx
xxolb
xxobl

In each of these cases only one sits next to another person, but not both. In every other conceivable case, both of them individually must sit next to at least one another person.

loxob
boxol
oblox
olbox
xolbo
xoblo
obolo
olobo
xoblo
xolbo

etc...
in 7 and 8 they are sitting next to 2 people and 9 and 10 is a repeat of 5 and 6.
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 08:28 PM   #10
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Quote:
Originally Posted by AW335TT View Post
in 7 and 8 they are sitting next to 2 people and 9 and 10 is a repeat of 5 and 6.
That's my point, except I didn't mean to repeat.
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 08:35 PM   #11
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

Quote:
Originally Posted by radix View Post
That's my point, except I didn't mean to repeat.
so that leaves only 6 possibilities
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 08:44 PM   #12
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

OK.... I got it, I just didn't interpret "one other person" to mean "one other specific person". When you read it that way, everything changes.


Let's call this one person Frank. The problem should have been more specific IMO, but for the sake of solving, I'll call him Frank. The probability that Frank takes a seat that is not on one of the ends is 3:5, or .60. The probability that the next person, say Bob, sits next to Frank is 2:4 (he can sit on either side) or .50. The probability that Linda takes the other seat next to Frank is 1:3, or .33 repeating. Therefore:

Code:
$ perl -le 'print 0.6 * 0.5 * 0.333333333'
0.0999999999
0.09999999999 = 10.

So the answer is 10%.
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 08:52 PM   #13
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

why wouldnt they want to sit on the ends? When they sit on the ends they are still sitting next to only 1 person.
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 09:01 PM   #14
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Quote:
Originally Posted by AW335TT View Post
why wouldnt they want to sit on the ends? When they sit on the ends they are still sitting next to only 1 person.
If Frank sat on the end, then Bob and Linda could not both sit next to him. The problem is badly worded, but in order to both sit next to this one person (hypothetically Frank), they must of course flank him or her on either side. This is not possible if this one person (Frank) sits in an end seat. The mistake the OP made is thinking that this problem should be solved using permutations. It shouldn't. It should be solved my multiplying the probabilities of successive events.
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 09:15 PM   #15
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

Are you sure that both Frank and Bob have to sit next to only one person (Linda) or do they individually have to sit next to other person.. like BOXOF or XOBFO. The question is veryy badly worded. So how would you get .90 for possibilities they wont only sit next to one person?
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 09:21 PM   #16
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Quote:
Originally Posted by AW335TT View Post
Are you sure that both Frank and Bob have to sit next to only one person (Linda) or do they individually have to sit next to other person.. like BOXOF or XOBFO. The question is veryy badly worded.

They have to sit like this:


xxbfl
xxlfb
xlfbx
xbflx
bflxx
lfbxx


In order for both Bob and Linda to sit next to this person (Frank), he must be in between them, otherwise only one of them is sitting next to him.
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 09:32 PM   #17
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

what my question was is that do both of them have to sit next to the same person or can they each sit next to 1 other person like LSxFB where both Linda and Bob are sitting next to one person
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 09:38 PM   #18
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

Quote:
Originally Posted by radix View Post
They have to sit like this:


xxbfl
xxlfb
xlfbx
xbflx
bflxx
lfbxx


In order for both Bob and Linda to sit next to this person (Frank), he must be in between them, otherwise only one of them is sitting next to him.
for #1 bob is sitting next to 2 people
2 Linda is sitting next to 2 people
3 Linda and bob are sitting next to 2 people same with 4
5 Linda is next to 2 people
6 Bob is next to 2 people
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 09:45 PM   #19
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Quote:
Originally Posted by AW335TT View Post
for #1 bob is sitting next to 2 people
2 Linda is sitting next to 2 people
3 Linda and bob are sitting next to 2 people same with 4
5 Linda is next to 2 people
6 Bob is next to 2 people
You're still misunderstanding the problem. It asks "what is the probability that both of them will sit next to only one other student from the group?". The only way they can both sit next to only one other student from the group, is if that student sits between them.
radix is offline   Philippines
0
Reply With Quote
      07-30-2010, 09:52 PM   #20
nightkhan
Private Woodman
 
nightkhan's Avatar
 
Drives: 328i, JB, Space Rocket
Join Date: Jan 2006
Location: San Francisco

Posts: 467
iTrader: (3)

slightly hijacking this thread temporarily, but i just took the gmat about a month ago, and i have to vouch for the manhattan gmat books. those books are amazing! helped me a lot and explained a lot of the little minute details and fundamentals that i never really understood 100%. definitely better than the official gmat books, and i took the kaplan class prob 2-3 years ago, and def blows that out of the water.

highly recommended
__________________
Former King of Stock...
E90 325i, Jet Black:
Done with
Current: E92 328i, Sports Package, 335i wheels, Power Seats, BMW Assist/Bluetooth, IPod/USB, heated seats, //M Short Shifter
Future Mods: 19" CSL's, KW V2, 15mm spacer, tints
nightkhan is offline   United_States
0
Reply With Quote
      07-30-2010, 10:17 PM   #21
AW335TT
Lieutenant
 
Drives: a car
Join Date: Feb 2007
Location: North Hollywood, CA

Posts: 476
iTrader: (0)

Quote:
Originally Posted by radix View Post
You're still misunderstanding the problem. It asks "what is the probability that both of them will sit next to only one other student from the group?". The only way they can both sit next to only one other student from the group, is if that student sits between them.
Ok but in the ones i mention one of them is not sitting next to only one other person he/she is sitting next to 2 people. Im going to ask my teacher tomorrow and see what he says.
__________________
AW335TT is offline   Armenia
0
Reply With Quote
      07-30-2010, 10:24 PM   #22
radix
there's something different about him
 
radix's Avatar
 
Drives: -
Join Date: Feb 2008
Location: -

Posts: 896
iTrader: (0)

Quote:
Originally Posted by AW335TT View Post
Ok but in the ones i mention one of them is not sitting next to only one other person he/she is sitting next to 2 people. Im going to ask my teacher tomorrow and see what he says.

Consider the problem this way. Instead of:

Quote:
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that both of them will sit next to only one other student from the group? The Answer is 10%
put names on all of them:

Quote:
Bob, Lisa, Frank, George, and Jen (five students) bought movie tickets in one row next to each other. What is the probability that Bob and Lisa will both sit next Frank (only one other student)? The Answer is 10%

Then the only possible positions that work with respect to Bob, Lisa, and Frank are:

jgbfl
jglfb
jlfbg
jbflg
bfljg
lfbjg

The positions of Jen and George don't matter with respect to probability. I don't know how else to explain it, maybe your teacher can do a better job.

Last edited by radix; 07-30-2010 at 10:30 PM.
radix is offline   Philippines
0
Reply With Quote
Post Reply

Bookmarks

Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump


All times are GMT -5. The time now is 05:37 AM.




m3post
Powered by vBulletin® Version 3.7.0
Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
1Addicts.com, BIMMERPOST.com, E90Post.com, F30Post.com, M3Post.com, ZPost.com, 5Post.com, 6Post.com, 7Post.com, XBimmers.com logo and trademark are properties of BIMMERPOST