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# Advanced Constraint Combinatorics

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Advanced Constraint Combinatorics [#permalink]  15 Feb 2007, 18:09
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

Last edited by GMAT100 on 15 Feb 2007, 21:11, edited 1 time in total.
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[#permalink]  15 Feb 2007, 18:51
6x5x4x3x2x1 = total number of arrangements = 720

total arrangements of J and M sitting together = 4x3x2x1x5x2 = 240

total arrangement of J and M not sitting together = 480

480
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[#permalink]  15 Feb 2007, 22:04
total ways of arranging 6 ppl = 6! = 6*5*4*3*2 = 720

ways in which m&j sit together = 5!*2 = 240

ans = 720 - 240 = 480
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[#permalink]  16 Feb 2007, 08:41
480 as well

total number of arrangements 6*5*4*3*2*1 = 720

want to count criteria we dont want

If Marcia sits in first chair, and Jan sits in 2nd chair there are 4! = 24 arrangements for the other people

Jan could also sit in first chair and Marcia in 2nd, so there are also 24 different arrangments

24+24 = 48

so we can have 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or 5th and 6th

therefore there are 5 ways each have 48 combinations so 48*5=240

720-240 = 480
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Re: Advanced Constraint Combinatorics [#permalink]  04 Aug 2011, 21:12
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.
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Re: Advanced Constraint Combinatorics [#permalink]  05 Aug 2011, 17:38
@Karishma - Can you please confirm the solutions and approach ?
I am posting 3 more variants of the same question.

Let me know whether the symmetrical approach can be applied here ?
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Re: Advanced Constraint Combinatorics [#permalink]  05 Aug 2011, 17:39
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 ?
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Re: Advanced Constraint Combinatorics [#permalink]  05 Aug 2011, 17:39
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 ?
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Re: Advanced Constraint Combinatorics [#permalink]  05 Aug 2011, 17:39
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 ?
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Re: Advanced Constraint Combinatorics [#permalink]  16 Aug 2011, 22:28
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 ?

-->

7! - 2((7-2+1)!) = 7!-2(6!)=7(6!)-2(6!) = 5(6!) ways

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.

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Re: Advanced Constraint Combinatorics [#permalink]  16 Aug 2011, 22:32
1
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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 ?

-->

8P7-2[(8-2+1)P(7-2+1)] = 8P7-2(7P6) ways = 8! - 2(7!) = 6(7!) ways

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!

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Re: Advanced Constraint Combinatorics [#permalink]  16 Aug 2011, 22:42
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.

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

-->

8P7-2[(8-2+1)P(7-2+1)] = 8P7-2(7P6) ways = 8! - 2(7!) = 6(7!) ways

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 be
8!-(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
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Re: Advanced Constraint Combinatorics [#permalink]  18 Aug 2011, 18:43
1
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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

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Re: Advanced Constraint Combinatorics [#permalink]  18 Aug 2011, 18:56
1
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.
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Re: Advanced Constraint Combinatorics [#permalink]  18 Aug 2011, 22:39
1
KUDOS
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.
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Re: Advanced Constraint Combinatorics [#permalink]  18 Aug 2011, 22:47
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.
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Re: Advanced Constraint Combinatorics [#permalink]  19 Aug 2011, 16:08
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 !!!
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Re: Advanced Constraint Combinatorics [#permalink]  19 Aug 2011, 16:12
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.
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Re: Advanced Constraint Combinatorics   [#permalink] 19 Aug 2011, 16:12
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