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# 4 people - sitting in a row seat

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Intern
Joined: 08 Nov 2009
Posts: 30
Location: United States
Concentration: Marketing, Entrepreneurship
GMAT 1: 720 Q49 V39
GPA: 3.4
WE: Engineering (Health Care)
Followers: 1

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

4 people - sitting in a row seat [#permalink]  18 Oct 2010, 12:40
00:00

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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Joined: 02 Sep 2010
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Kudos [?]: 299 [1] , given: 25

Re: 4 people - sitting in a row seat [#permalink]  18 Oct 2010, 14:18
1
KUDOS
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)

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

This is a tough question, and the lieklihood of getting it on the GMAT is very low
_________________
Intern
Joined: 08 Nov 2009
Posts: 30
Location: United States
Concentration: Marketing, Entrepreneurship
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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Joined: 02 Sep 2009
Posts: 11515
Followers: 1791

Kudos [?]: 9533 [0], given: 826

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

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

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