GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 19 Feb 2020, 20:59

### 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

# 6 persons are going to theater and will sit next to each oth

Author Message
TAGS:

### Hide Tags

Intern
Joined: 16 Jun 2010
Posts: 2
6 persons are going to theater and will sit next to each oth  [#permalink]

### Show Tags

16 Jun 2010, 19:08
00:00

Difficulty:

25% (medium)

Question Stats:

88% (01:20) correct 12% (01:40) wrong based on 19 sessions

### HideShow timer Statistics

6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !

It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives

I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why

regards
SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1813
Concentration: General Management, Nonprofit

### Show Tags

16 Jun 2010, 19:45
1
Total number of arrangements = 6!

Assuming the two sit next to each other, we have 5!x2 arrangements (This is because when they are sitting next to each other, we can consider both of them as one "unit" and hence there are 5 units, i.e them and the other 4 people. This will lead to an arrangement of 5! and then between them, they can be seated in two ways, so it's 2x5!)

So answer = Total - Arrangements with them sitting next to each other

= 6! - 2x5! = 5! x 4 = 480

I think you assumed they were sitting next to each other and did only the 2! and 4!

What are the numbers given in the official answer? I believe you only mentioned word choices.
Math Expert
Joined: 02 Sep 2009
Posts: 61302

### Show Tags

16 Jun 2010, 19:54
1
1
Maude wrote:
Hello

Please need help to understand the following

6 persons are going to theater and will sit next to each other in 6 adjacent seats
But Martia and Jan can not sit next to each other .In how many Arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !

It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives

I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why

regards

Hi, and welcome to Gmat Club! Below is the solution for your problem.

You are right saying that probably the best way to deal with the questions like this is to count total # of arrangements and then subtract # of arrangements for which opposite of restriction occur. But the way you are calculating the later is not correct.

Total # of arrangements of 6 people (let's say A, B, C, D, E, F) is $$6!$$.
# of arrangement for which 2 particular persons (let's say A and B) are adjacent can be calculated as follows: consider these two persons as one unit like {AB}. We would have total 5 units: {AB}{C}{D}{E}{F} - # of arrangement of them 5!, # of arrangements of A and B within their unit is 2!, hence total # of arrangement when A and B are adjacent is $$5!*2!$$.

# of arrangement when A and B are not adjacent is $$6!-5!*2!$$.

Total # of arrangements of 5 distinct digits is $$5!$$.
# of arrangement for which 2 digits 2 and 4 are adjacent is: consider these two digits as one unit like {24}. We would have total 4 units: {24}{1}{3}{5} - # of arrangement of them 4!, # of arrangements of 2 and 4 within their unit is 2!, hence total # of arrangement when 2 and 4 are adjacent is $$4!*2!$$.

# of arrangement when 2 and 4 are not adjacent is $$5!-4!*2!$$.

Hope it helps.
_________________
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3155
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

### Show Tags

22 Dec 2015, 01:12
Maude wrote:
6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

6 person can sit on 6 seats in way = 6*5*4*3*21 = 6! = 720

6 person can sit on 6 seats in ways such that Martia and Jan sit next to each other = (5*4*3*2*1)*(2!) = 5!*2! = 240

Favorable cases = 720 - 240 = 480
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Senior Manager
Joined: 15 Oct 2015
Posts: 297
Concentration: Finance, Strategy
GPA: 3.93
WE: Account Management (Education)
Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

### Show Tags

22 Dec 2015, 02:40
the quick formular for this kind of questions is solve removing the constraint - the opposite of the constraint.
Hence, 6! - 5!
the opposite of the constraint is if the two guys must sit next to each other, that is, they are one, hence you have 5.
how many ways can you arrange 6 unique letters minus how many ways can you arrange 5 unique letters
Math Expert
Joined: 02 Aug 2009
Posts: 8256
Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

### Show Tags

22 Dec 2015, 03:06
Nez wrote:
the quick formular for this kind of questions is solve removing the constraint - the opposite of the constraint.
Hence, 6! - 5!
the opposite of the constraint is if the two guys must sit next to each other, that is, they are one, hence you have 5.
how many ways can you arrange 6 unique letters minus how many ways can you arrange 5 unique letters

Hi,
you have found the correct method but missed out the final step..
both are taken as one and we get the arrangement as 5! but within the two both can be arranged in 2! ways..
_________________
Director
Joined: 27 May 2012
Posts: 952
Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

### Show Tags

07 Sep 2018, 10:01
Maude wrote:
6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !

It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives

I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why

regards

Dear Moderator,
How can it be A , when no choices are given, hope you will look into this. Thank you.
_________________
- Stne
Manager
Joined: 18 Jun 2018
Posts: 249
Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

### Show Tags

07 Sep 2018, 10:07
Maude wrote:
6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !

It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives

I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why

regards

Total number of arranging 6 persons $$=6!$$
Total number of arranging 6 persons such that Matina and Jan are together $$=2*5!$$
the number of arranging 6 persons such that Matina and Jan are not together $$= 6!-2*5!$$
Re: 6 persons are going to theater and will sit next to each oth   [#permalink] 07 Sep 2018, 10:07
Display posts from previous: Sort by