GMPJBC go to movie and they sit in adjacent row of 6 seats. : GMAT Problem Solving (PS)
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# GMPJBC go to movie and they sit in adjacent row of 6 seats.

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GMPJBC go to movie and they sit in adjacent row of 6 seats.  [#permalink]

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26 Dec 2007, 13:10
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75% (02:23) correct 25% (02:16) wrong based on 6 sessions

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GMPJBC go to movie and they sit in adjacent row of 6 seats. But, MJ cannot sit next to each other. How many different arrangements are possible?

Can someone explain why my answer is wrong?

My way:
Approach: figure total number of arrangements and then subtract from this total, the number of MJ arrangements

1. total ways: 6!
2. total MJ/JM ways = 10 ways (i drew it out, beginning with seat, 12,23,34,45,56)

I know that there is something wrong with #2 because the answer is 720-240 = 480.

Can someone explain how they got 240?

Source: MGMAT.
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26 Dec 2007, 13:17
aliensoybean wrote:
GMPJBC go to movie and they sit in adjacent row of 6 seats. But, MJ cannot sit next to each other. How many different arrangements are possible?

Can someone explain why my answer is wrong?

My way:
Approach: figure total number of arrangements and then subtract from this total, the number of MJ arrangements

1. total ways: 6!
2. total MJ/JM ways = 10 ways (i drew it out, beginning with seat, 12,23,34,45,56)

I know that there is something wrong with #2 because the answer is 720-240 = 480.

Can someone explain how they got 240?

Source: MGMAT.

this is an easy question. u simply didnt understand the concept.

it is 1- P(they DO sit together)

6! - 5! 2! = 480
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26 Dec 2007, 13:19
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u have 6 ppl. 2 cannot sit together. so it is 1-p(they do sit together)

p(they do sit together) implies that we group those 2 guys into 1 set. This means we have 6-2+1= 5 people to arrange. We also have to account for the arrangement within the set because AB is different from BA. So it is 5!2!

http://www.manhattangmat.com/strategyseries11.cfm
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28 Jul 2009, 15:47
Hi!

I found a pretty similar problem in the following book: "Probability for dummies" - just look it up on Google Books and search for the following string within the book: "Soma". You will find a very similar problem that is solved using aliensoybean's approach.

So my question is why are the posts in this forum right and the book wrong (or maybe both are right but please explain to me

Thanks!
Steve
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19 Sep 2009, 07:41
enfinity wrote:
Hi!

I found a pretty similar problem in the following book: "Probability for dummies" - just look it up on Google Books and search for the following string within the book: "Soma". You will find a very similar problem that is solved using aliensoybean's approach.

So my question is why are the posts in this forum right and the book wrong (or maybe both are right but please explain to me

Thanks!
Steve

Steve,

The approach presented above is correct.

If you want, post here the similar problem you mentioned, and I'll have a go at it.

Cheers
_________________

Please kudos if my post helps.

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19 Sep 2009, 08:00
Thanks man! Here it comes:

You have four friends: Jim, Arun, Soma, and Eric. How many ways can you rearrange the individuals in a row so that Soma and Eric do not sit next to each other.

My approach (and the one presented in the book i've mentioned):

Find all possibilities: 4!=24 (certain scenarios are not possible; therefore, we need to subtract these impossibilities):

Find the ways in which S and E sit next to each other:

E S _ _
_ E S _
_ _ E S

OR

S E _ _
_ S E _
_ _ S E

Therefore, 24 - 6 = 18
Manager
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19 Sep 2009, 08:31
enfinity wrote:
Thanks man! Here it comes:

You have four friends: Jim, Arun, Soma, and Eric. How many ways can you rearrange the individuals in a row so that Soma and Eric do not sit next to each other.

My approach (and the one presented in the book i've mentioned):

Find all possibilities: 4!=24 (certain scenarios are not possible; therefore, we need to subtract these impossibilities):

Find the ways in which S and E sit next to each other:

E S _ _
_ E S _
_ _ E S

OR

S E _ _
_ S E _
_ _ S E

Therefore, 24 - 6 = 18

This approach is similar to the other one, but it has a mistake.

1st. All possibilities = 4! = 24
2nd. Take ES as 1 person. So now we have 3 people: J, A & ES --> 3! = 6
3nd. Finally, it is necessary to account for different arrangements within ES --> 2! = 2
=> All - P (they DO sit together) = 24 - 3! * 2! = 24 - 12 = 12

By multiplying by 3 instead of 3! you are not accounting for the different positions of J and A. Have a look at the following:

S E J A
J S E A
J A S E

Is not the same as

S E A J
A S E J
A J S E

But the approach presented on that book does not take that into account.

Back to the original problem:
=> All - P (they DO sit together) = 6! - 5! * 2! = 720 - 120 * 2 = 480

Hope that helps.

Also, please know that probability is not my forte.
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27 Sep 2009, 10:58
GMPJBC go to movie and they sit in adjacent row of 6 seats. But, MJ cannot sit next to each other. How many different arrangements are possible?

Soln:
Arrangements in which MJ sit together is = 5! * 2!
Total Arrangemets possible with 6 people is = 6!

Arrangements in which MJ dont sit together is
= Total arrangements - Arrangements where they sit together
= 6! - 5! * 2!
= 480 ways
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07 Feb 2010, 21:56
nice one.

1 - p(when they sit together) , the manhattan link was really helpful. thx
Re: Combo/ Permu   [#permalink] 07 Feb 2010, 21:56
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# GMPJBC go to movie and they sit in adjacent row of 6 seats.

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