[Makushr1]

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.

[AW335TT]

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?

[Makushr1]

Originally Posted by AW335TT

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...

[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.

[radix]

Are you sure the question doesn't specify how many seats are in a row, and that the number isn't more than five?

[AW335TT]

So isn't 6 10% of 60? 120 possibilities divided by 2 ppl. Why did you do 5!/(3!2!)?

[AW335TT]

Originally Posted by radix

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.

[radix]

Originally Posted by AW335TT

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...

[AW335TT]

Originally Posted by radix

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.

[radix]

Originally Posted by AW335TT

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.

[AW335TT]

Originally Posted by radix

That's my point, except I didn't mean to repeat.

so that leaves only 6 possibilities

[radix]

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:

$ perl -le 'print 0.6 * 0.5 * 0.333333333'
0.0999999999

0.09999999999 = 10.

So the answer is 10%.

[AW335TT]

why wouldnt they want to sit on the ends? When they sit on the ends they are still sitting next to only 1 person.

[radix]

Originally Posted by AW335TT

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.

[AW335TT]

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?

[radix]

Originally Posted by AW335TT

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.

[AW335TT]

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]

Originally Posted by radix

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

[radix]

Originally Posted by AW335TT

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.

[nightkhan]

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

[AW335TT]

Originally Posted by radix

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.

[radix]

Originally Posted by AW335TT

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:

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:

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.