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

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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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

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

Hope it helps.
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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.

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?

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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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.

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.

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?

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

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.

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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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)

Simple derangement question.

Deragement(4)/ 4!

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