4 people - sitting in a row seat

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4 people - sitting in a row seat [#permalink]

18 Oct 2010, 12:40

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Question Stats:

40% (02:36) correct

60% (00:16) wrong

based on 0 sessions

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up.
When they return, they each randomly sit down on one of the 4 chairs.
What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

a. 3/24
b. 9/24
c. 15/24
d. 18/24
e. 21/24

(I am getting a different result than the OA - a result that even does not appear in the answer choices. Please collaborate)

[Reveal] Spoiler: OA

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Re: 4 people - sitting in a row seat [#permalink]

18 Oct 2010, 14:18

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eladshus wrote:

Ways all 4 sit in the correct seat = 1
Ways such that exactly 3 in correct seat = 0 (If there is one person in the wrong seat, then whose seat it is must also be in the wrong seat)
Ways such that exactly 2 in wrong seat = C(4,2) = 6 (choose the two in wrong seat, only one way to get it wrong)
Ways such that exactly 3 in wrong seat = C(4,1) x (3! - C(3,2) - 1) = 4x2= 8 (choose the one in the correct seat x the ways 3 guys can be in 3 wrong seats -- which is again all ways to seat 3 people - ways in which all correct (1) - ways in which 2 wrong (C(3,2)) )

So total ways to get all 4 wrong = 4! - 1 - 0 - 6 - 8 = 9

So probability = 9/24
Answer : (b)

This is a tough question, and the lieklihood of getting it on the GMAT is very low
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Intern

Joined: 08 Nov 2009

Posts: 30

Location: United States

Concentration: Marketing, Entrepreneurship

Schools: Booth '14 (I), Sloan '14, Ross '14 (D), Haas '14 (S), Fuqua '14 (I)

GMAT 1: 720 Q49 V39

GPA: 3.4

WE: Engineering (Health Care)

Followers: 1

Kudos [?]: 10 [0], given: 9

Re: 4 people - sitting in a row seat [#permalink]

18 Oct 2010, 14:47

Man - you played it!
Thanks. I didn't think of all the possibilities.

My initial solution was:

3/4 probability for the first person to NOT sit on his chair
2/3 probability for the second person to NOT sit on his chair (from the remaining 3 chairs)
1/2 probability for the third person to NOT sit on his chair (from the remaining 2 chairs)

So, my final solution was: 3/4 * 2/3 * 1/2 = 1/4

But I missed some arrangements.

Thanks

+1 from me

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Re: 4 people - sitting in a row seat [#permalink]

19 Oct 2010, 12:55

eladshus wrote:

All other possible scenarios discussed here: letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letter%20arrangements

Hope it helps.
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Re: 4 people - sitting in a row seat [#permalink] 19 Oct 2010, 12:55

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4 people - sitting in a row seat

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4 people - sitting in a row seat

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