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A group of 5 students bought movie tickets in one row next

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A group of 5 students bought movie tickets in one row next [#permalink]

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New post 12 Aug 2013, 01:40
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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 Bob and Lisa will each sit next to only one of the four other students from the group?

(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%

Can somebody plz clarify that the answer to this prob is 4!*2 which I am getting, but the answer in the book is given something else.
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Aug 2013, 02:05, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: A group of 5 students bought movie tickets in one row next [#permalink]

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New post 12 Aug 2013, 02:05
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Countdown wrote:
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 Bob and Lisa will each sit next to only one of the four other students from the group?

(A)5%
(B)10%
(C)15%
(D)20%
(E)25%

Can somebody plz clarify that the answer to this prob is 4!*2 which I am getting, but the answer in the book is given something else.



The question basically asks about the probability that Bob and Lisa sit at the ends.

The total # of sitting arrangements is 5!.

Desired arrangement is either BXYZL or LXYZB. Now, XYZ can be arranged in 3! ways, therefore total # of favorable arrangements is 2*3!.

P=(favorable)/(total)=(2*3!)/5!=1/10.

Answer: B.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3, 7, and 8. Thank you.
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Re: Permutation [#permalink]

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New post 12 Aug 2013, 02:05
Countdown wrote:
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 number of ways of seating if both of them will sit next to only one other student from the group?



Can somebody plz clarify that the answer to this prob is 4!*2 which I am getting, but the answer in the book is given something else.


LET SAY THAT BOB = B ..LISA = L
LET ONE OTHER GUY = X
NOW see question is saying....B AND L will sit next to this GUY only I.E X.
what you have done is you assumed that question is asking that B and L will sit together.
so according to given condition X cannot sit at the corner seat because in that case both B and L cant sit next to X.
so FOR X we can make hime arrange in 3!ways and B and L can be arranged in 2! ways.
therefore total ways = \(3!*2!\)

hope it helps.
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Re: A group of 5 students bought movie tickets in one row next [#permalink]

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New post 12 Aug 2013, 09:53
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Countdown wrote:
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 Bob and Lisa will each sit next to only one of the four other students from the group?

(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%

Can somebody plz clarify that the answer to this prob is 4!*2 which I am getting, but the answer in the book is given something else.

use this diagram : it can help
Attachments

bob lisa.png
bob lisa.png [ 24.68 KiB | Viewed 2894 times ]


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Re: A group of 5 students bought movie tickets in one row next [#permalink]

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Re: A group of 5 students bought movie tickets in one row next [#permalink]

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New post 26 Mar 2017, 01:37
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A group of 5 students bought movie tickets in one row next   [#permalink] 26 Mar 2017, 01:37
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