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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?

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 Answer : (b)

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

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)

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)

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.

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);

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);

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

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09 Oct 2013, 02:35

Expert's post

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);

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

Bunuel, Can you Please comment?

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.

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.

eladshus wrote:

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)

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

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28 Oct 2013, 18:17

eladshus 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.

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

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?

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

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. _________________

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:

eladshus 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.

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

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02 Jun 2014, 09:35

Expert's post

jlgdr wrote:

saurabhprashar wrote:

eladshus 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.

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

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02 Jun 2014, 16:59

Expert's post

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) _________________

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

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