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# 4 people are sitting in a 4 seat row watching a football

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4 people are sitting in a 4 seat row watching a football [#permalink]

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18 Oct 2010, 12:40
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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
[Reveal] Spoiler: OA

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

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

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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

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

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

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25 Jul 2013, 04:40
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

hi....I also tried this method but later realized that we are making the probability of the last one to choose wrong seat as 1 in this method. (3/4 * 2/3 * 1/2 *1) But that is not true.

The first 3 guys can interchange their seats, all sit in wrong seats and the last one can still sit in his own place. We are missing something here.

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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25 Jul 2013, 15:47
You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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08 Oct 2013, 23:47
R0dman wrote:
You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

I am getting (9+6)/24 as answer?

Which is equal to
15/24 = 5/8

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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09 Oct 2013, 02:35
honchos wrote:
R0dman wrote:
You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

I am getting (9+6)/24 as answer?

Which is equal to
15/24 = 5/8

Please check here: 4-people-are-sitting-in-a-4-seat-row-watching-a-football-103089.html#p802345 and here: 4-people-are-sitting-in-a-4-seat-row-watching-a-football-103089.html#p803019
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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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09 Oct 2013, 06:44
honchos wrote:
R0dman wrote:
You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

I am getting (9+6)/24 as answer?

Which is equal to
15/24 = 5/8

Once again. it's "1- negative outcome". Probability of negative outcome is in brackets. So it goes (1/4+1/4+1/8) = (6/24+6/24+3/24) =15/24.
Then you deduct it from 1.
So basically you getting 1- 15/24 = 9/24.

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

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10 Oct 2013, 23:53
I dont know whether this is the right way to approach the problem , however i get the same ans.

Originally seated A B C D

now when after they get up and when they sit back again .
1st- A has option to sit on 3 seats ( apart from his previous seat . thus he now sits on B's seat.)
2nd- Similarly B has option to sit on 3 seats ( because A has already occupied B's previous seat, thus B sits on a's seat.)
3rd- Now C has only 1 option to sit on D's seat . and similarly D also has one option to sit on C's seat.)

hence total favourable outcomes 3*3*1*1=9

and total possible outcomes =4!=24

probability of the favourable outcome= 9/24.

Please correct me if i am wrong.

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 are sitting in a 4 seat row watching a football [#permalink]

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28 Oct 2013, 18:17
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

There is a formula for total number of possible de arrangements:

Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own envelope is = n! [ (1/2!) - (1/3!) + (1/4!) ...... (-1)^n (1/n!) ]

In this case n=4. Therefore, total number of possible dearrangements such that none of them sit in their place = 4! [ (1/2!) - (1/3!) + (1/4!) ] = 9.

Likelihood = Favorable outcomes/Total outcomes = 9/4! = 9/24.

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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31 Oct 2013, 07:17
Hi guys, this is my first post.
Here is my way to the solution:
when the 4 people come back the first one has 3/4 possibilities to choose a seat different from the previous one, let's call him A and assume he seats were B used to seat before;
then B for sure will be sitting on a different place, because A is now on his place, so B's possibility to choose a different seat is 1;
C will have 2 choices, and just one suitable for sitting on a different seat, so C's possibility will be 1/2;
D will have no choices, because C is now on his seat, so also D will have possibility 1 to sit on a different seat.
3/4 * 1 * 1/2 * 1 = 3/8 that is the same of 9/24

Is there any flaw in the above?

Thanks a lot

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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31 Oct 2013, 10:02
bugatti wrote:
Hi guys, this is my first post.
Here is my way to the solution:
when the 4 people come back the first one has 3/4 possibilities to choose a seat different from the previous one, let's call him A and assume he seats were B used to seat before;
then B for sure will be sitting on a different place, because A is now on his place, so B's possibility to choose a different seat is 1;
C will have 2 choices, and just one suitable for sitting on a different seat, so C's possibility will be 1/2;
D will have no choices, because C is now on his seat, so also D will have possibility 1 to sit on a different seat.
3/4 * 1 * 1/2 * 1 = 3/8 that is the same of 9/24

Is there any flaw in the above?

Thanks a lot

let's call him A and assume he seats were B used to seat before;, what if A doent sit where B sat?

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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31 Oct 2013, 17:28
Quote:
what if A doent sit where B sat?

Yes you are right, that's the flaw. By changing the choice made by A at the beginning changes the probability as a whole. For instance if A sits where C used to seat, then B will have 2/3 of possibilities to sit on a different seat. It doesn't work

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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17 Nov 2013, 09:39
I got the correct answer, its simple, this problem is based on the Derangement Formula. Search the net and read about it, it comes to use in many problem scenarios.
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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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02 Jun 2014, 09:20
saurabhprashar wrote:
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

There is a formula for total number of possible de arrangements:

Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own envelope is = n! [ (1/2!) - (1/3!) + (1/4!) ...... (-1)^n (1/n!) ]

In this case n=4. Therefore, total number of possible dearrangements such that none of them sit in their place = 4! [ (1/2!) - (1/3!) + (1/4!) ] = 9.

Likelihood = Favorable outcomes/Total outcomes = 9/4! = 9/24.

Bunuel, on dearrangement formula, should one learn this for GMAT?

Cheers!
J

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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02 Jun 2014, 09:35
jlgdr wrote:
saurabhprashar wrote:
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

There is a formula for total number of possible de arrangements:

Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own envelope is = n! [ (1/2!) - (1/3!) + (1/4!) ...... (-1)^n (1/n!) ]

In this case n=4. Therefore, total number of possible dearrangements such that none of them sit in their place = 4! [ (1/2!) - (1/3!) + (1/4!) ] = 9.

Likelihood = Favorable outcomes/Total outcomes = 9/4! = 9/24.

Bunuel, on dearrangement formula, should one learn this for GMAT?

Cheers!
J

No. This is out of the scope of the GMAT.
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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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02 Jun 2014, 16:59
killer question..

X - Y - Z - V <--- Original Arrangement (below are the potential seats each person can sit in to satisfy constraint)
* - X - X - X
Y - * - Y - Y
Z - Z - * - Z
V - V - V - *

starting from the first non-underlined X, you can pick any combo of the below letters a long as they don't occupy the same column (i.e. seat). there are 3 such arrangements for each X or 9 total (divided by 4! or 24 total possibilities)
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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

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20 Jul 2017, 07:33
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

Simple derangement question.

Deragement(4)/ 4!
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Re: 4 people are sitting in a 4 seat row watching a football   [#permalink] 20 Jul 2017, 07:33
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