It is currently 18 Oct 2017, 04:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 11 Nov 2007
Posts: 60

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

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

### Show Tags

26 Dec 2007, 14:10
2
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

75% (01:23) correct 25% (02:16) wrong based on 8 sessions

### HideShow timer Statistics

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.

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

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1045 [0], given: 4

Location: New York City

### Show Tags

26 Dec 2007, 14:17
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

Kudos [?]: 1045 [0], given: 4

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1045 [1], given: 4

Location: New York City

### Show Tags

26 Dec 2007, 14:19
1
KUDOS
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

Kudos [?]: 1045 [1], given: 4

Manager
Joined: 12 May 2009
Posts: 53

Kudos [?]: 117 [0], given: 18

### Show Tags

28 Jul 2009, 16:47
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

Kudos [?]: 117 [0], given: 18

Manager
Joined: 22 Jul 2009
Posts: 191

Kudos [?]: 281 [0], given: 18

### Show Tags

19 Sep 2009, 08:41
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.

Kudos [?]: 281 [0], given: 18

Manager
Joined: 12 May 2009
Posts: 53

Kudos [?]: 117 [0], given: 18

### Show Tags

19 Sep 2009, 09:00
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

Kudos [?]: 117 [0], given: 18

Manager
Joined: 22 Jul 2009
Posts: 191

Kudos [?]: 281 [0], given: 18

### Show Tags

19 Sep 2009, 09:31
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.
_________________

Please kudos if my post helps.

Kudos [?]: 281 [0], given: 18

Manager
Joined: 27 Oct 2008
Posts: 185

Kudos [?]: 164 [0], given: 3

### Show Tags

27 Sep 2009, 11:58
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

Kudos [?]: 164 [0], given: 3

Intern
Joined: 11 Jul 2009
Posts: 8

Kudos [?]: [0], given: 19

### Show Tags

07 Feb 2010, 22:56
nice one.

1 - p(when they sit together) , the manhattan link was really helpful. thx

Kudos [?]: [0], given: 19

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16713

Kudos [?]: 273 [0], given: 0

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

### Show Tags

07 Sep 2017, 10:16
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Re: GMPJBC go to movie and they sit in adjacent row of 6 seats.   [#permalink] 07 Sep 2017, 10:16
Display posts from previous: Sort by