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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and [#permalink]
15 Feb 2007, 17:09

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

80% (01:32) correct
20% (00:00) wrong based on 0 sessions

Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements

Last edited by GMAT100 on 15 Feb 2007, 20:11, edited 1 time in total.

Re: Advanced Constraint Combinatorics [#permalink]
04 Aug 2011, 20:12

2

This post received KUDOS

Expert's post

GMAT100 wrote:

The following Question:

Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If Marcia and Jan will not sit next to each other, in how many different arrangements

Answer will be posted tomorrow

@krishp84

krishp84 wrote:

why cannot we apply symmetry here ?

We cannot apply symmetry because the number of cases where M and J are sitting together is not equal to the number of cases where M and J are not sitting together. Consider 3 people: A, B and C They can be arranged in 3! = 6 ways ABC ACB BAC BCA CAB CBA In how many of these are A and B sitting together? 2 In how many are they not sitting together? 4 When we arrange 6 people here in 6! (= 720) ways, in how many of those will M and J sit together? Consider M and J to be one and arrange 5 people in 5! ways. Also, M and J can exchange places so multiply by 2. You get 5!*2 = 240 ways Out of 720, M and J will be next to each other in only 240 ways.

krishp84 wrote:

I recall a question related to this - "6 people standing in a line and one person(A) cannot stand behind another person(B). What is the total number of ways of forming the line? " Cannot remember the complete question, something like this...But remember that it was pure symmetry : 6!/2=360 ways

That question would be something like this: 6 people go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If Marcia will not sit to the right of Jan, how many different arrangements are possible?

'to the right of Jan' means anywhere on the right, not necessarily on the adjacent seat. Here we see symmetry because there are only 2 ways in which Marcia can sit. She can sit either to the left of Jan (any seat on the left) or to the right of Jan (any seat on the right). There is nothing else possible. The number of cases in which she will sit to the left will be same as the number of cases in which she will sit to the right. That is why the answer here will be 6!/2.

But the original question talks about sitting right next to each other on adjacent seats. The probability of sitting 'next to each other' is less than the probability of sitting 'not next to each other'. Hence we cannot apply the symmetry principle there.

Re: Advanced Constraint Combinatorics [#permalink]
05 Aug 2011, 16:39

Variant-2.1 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?

Re: Advanced Constraint Combinatorics [#permalink]
05 Aug 2011, 16:39

Variant-3.1 of this question -

7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ?

Re: Advanced Constraint Combinatorics [#permalink]
05 Aug 2011, 16:39

Variant-4.1 of this question -

7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?

Re: Advanced Constraint Combinatorics [#permalink]
16 Aug 2011, 21:28

Expert's post

Variant-1 of this question - 7 people(A,B,C,D,E,F.H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ?

Yes, that's fine. You make them sit together and subtract that out of 7!.

Variant-2 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A,F and E will not sit next to each other in how many different arrangements ?

7!-(3!)(7-3+1)! = 7!- 6(5!) ways

A little ambiguity here. I would say - A, F and E will not all sit together in how many ways? The solution is fine.

Re: Advanced Constraint Combinatorics [#permalink]
16 Aug 2011, 21:32

1

This post received KUDOS

Expert's post

Variant-3 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ?

The answer is correct. An alternative approach could be this:

I will just think of the vacant seat as a person V. So 8 people, 8 seats in 8! ways. 2 people should not be together so 8! - 2*7!

Another interesting variant could be this: 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?

Try it.

Variant-4 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A ,F and E will not sit next to each other in how many different arrangements ?

Correct solution. Alternative approach parallel to the one above:

The vacant spot is a person V. Required arrangements: 8! - 3!*6!

Re: Advanced Constraint Combinatorics [#permalink]
16 Aug 2011, 21:42

Expert's post

Variant-1.1 of this question - 7 people(A,B,C,D,E,F.H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ?

--> 7!/2 ways

Correct.

Variant-2.1 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?

Variant-3.1 of this question -

7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ?

Variant-4.1 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?

You can apply symmetry in all these questions. Tell me how you will do it. I will confirm the answer.

Re: Advanced Constraint Combinatorics [#permalink]
18 Aug 2011, 17:30

VeritasPrepKarishma wrote:

Variant-3 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ?

The answer is correct. An alternative approach could be this:

I will just think of the vacant seat as a person V. So 8 people, 8 seats in 8! ways. 2 people should not be together so 8! - 2*7!

WoW - Loved this Approach --> Kudos +1

VeritasPrepKarishma wrote:

Another interesting variant could be this: 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?

Try it.

This is equivalent to saying 8 people sit in 8 adjacent seats with an imaginary girl Ms. X with them Subtract the number of arrangements when Ms. X sits between A and F AXF FXA XAF XFA AFX FAX All the above are equivalent - No person sits between A and F So answer will be8!-(3!)(6!) ways

Now let me apply my traditional approach and confirm this - 8P7-(3!)((8-3+1)P(7-3+1)) = 8P7-(3!)(6P5) = 8!-(3!)(6!) ways

Re: Advanced Constraint Combinatorics [#permalink]
18 Aug 2011, 17:43

1

This post received KUDOS

krishp84 wrote:

Variant-2.1 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?

Let me post my solution - This will be equivalent as below -

A can come to left/right of F F can come to left/right of E (I diagrammed, but it is time-taking to create a picture and post here)

So number of different arrangements = (1/2)(7!/2) = 7!/4 ways

Re: Advanced Constraint Combinatorics [#permalink]
18 Aug 2011, 17:56

1

This post received KUDOS

VeritasPrepKarishma wrote:

Variant-3.1 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ?

This will be 8!/2 ways or (8P7)/2 = 8!/2 ways

VeritasPrepKarishma wrote:

Variant-4.1 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?

This will be (1/2)(8!/2) = 8!/4 ways or (1/2)(1/2)(8P7) = 8!/4 ways Logic - combination of variant 2.1 and 3.1

VeritasPrepKarishma wrote:

You can apply symmetry in all these questions. Tell me how you will do it. I will confirm the answer.

I understand this is a HUGE post - But could not find another way to link all these related concepts. Welcome to any new variant. I LOVE these puzzles.

Re: Advanced Constraint Combinatorics [#permalink]
18 Aug 2011, 21:39

1

This post received KUDOS

Expert's post

VeritasPrepKarishma wrote:

Variant-3 of this question - 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ?

-->

I will just think of the vacant seat as a person V. So 8 people, 8 seats in 8! ways. 2 people should not be together so 8! - 2*7!

Another interesting variant could be this: 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?

krishp84 wrote:

This is equivalent to saying 8 people sit in 8 adjacent seats with an imaginary girl Ms. X with them Subtract the number of arrangements when Ms. X sits between A and F AXF FXA XAF XFA AFX FAX All the above are equivalent - No person sits between A and F So answer will be 8!-(3!)(6!) ways

This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and also those cases where A and F have V (or your Ms. X!) between them are not acceptable. Number of cases where A and F are together = 2*7! Number of cases where A and F have V between them = 2*6! [AVF and FVA] These are the cases that are not acceptable. Answer should be 8! - 2*7! - 2*6! = 8! - 16*6!

Think about it another way: Compare the two questions. One where you don't want them to be together, the other where you don't want them to be together and you don't want the vacant spot between them. Obviously, in the second case, the number of cases you do not want are higher. In the first question, you subtracted a total of 2*7! arrangements i.e. 14*6! arrangements. In the second question, you need to subtract some more. You cannot subtract 3!*6! = 6*6! arrangements only.

Re: Advanced Constraint Combinatorics [#permalink]
18 Aug 2011, 21:47

Expert's post

krishp84 wrote:

I understand this is a HUGE post - But could not find another way to link all these related concepts.[/color] Welcome to any new variant. I LOVE these puzzles.

Your solutions are correct. In a few weeks, I am going to start Permutation Combination on my blog. I already intend to make a post on this question and all its variants. If you want to try out further complications, try putting in 2 or 3 vacant spots. Now there will be 2-3 identical people named 'V' so adjustments will be needed.

Re: Advanced Constraint Combinatorics [#permalink]
19 Aug 2011, 15:08

VeritasPrepKarishma wrote:

This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and also those cases where A and F have V (or your Ms. X!) between them are not acceptable. Number of cases where A and F are together = 2*7! Number of cases where A and F have V between them = 2*6! [AVF and FVA] These are the cases that are not acceptable. Answer should be 8! - 2*7! - 2*6! = 8! - 16*6!

Think about it another way: Compare the two questions. One where you don't want them to be together, the other where you don't want them to be together and you don't want the vacant spot between them. Obviously, in the second case, the number of cases you do not want are higher. In the first question, you subtracted a total of 2*7! arrangements i.e. 14*6! arrangements. In the second question, you need to subtract some more. You cannot subtract 3!*6! = 6*6! arrangements only.

Yes - I made that mistake of assuming X/V as a part of the group with A,F instead of thinking them separately. Stupid me !!!

Re: Advanced Constraint Combinatorics [#permalink]
19 Aug 2011, 15:12

VeritasPrepKarishma wrote:

If you want to try out further complications, try putting in 2 or 3 vacant spots. Now there will be 2-3 identical people named 'V' so adjustments will be needed.

Yes - I was thinking on these lines....I will post some variants on this once I am free.

Also thought of combining Probability with this because it is so related especially - the symmetry part.

Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and [#permalink]
15 Dec 2013, 03:47

1

This post received KUDOS

Expert's post

GMAT100 wrote:

Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements

The best way to deal with the questions like this is to find the total # of arrangements and then subtract # of arrangements for which opposite of restriction occur:

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!=480.