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Question on combinations and permutations

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Question on combinations and permutations [#permalink] New post 11 Nov 2009, 09:27
So if you have 6 people going to a movie and 6 seats next to each other for them to sit, but 2 of the people, Ron and Todd, will not sit next to each other you have 6! - (5! + 5!) ways to sit everybody in this mannor. This is the total number of ways to sit 6 people, minus each of the possible ways that the two people, Ron and Todd, can sit next to each other.

First question: Is there a better strategy to solve this type of question without addition/subtraction?

Now although this may or may not be something the GMAT will ask, I am curious to the following. Lets say that of the 6 people, the two that wont sit next to each other, Ron and Todd, have decided that they WILL sit with EXACTLY 1 person in between them. How many different ways can the 6 people sit while having Ron and Todd separated by exactly 1 person? :shock:

The math/strategy behind the second part would really be appreciated.
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Re: Question on combinations and permutations [#permalink] New post 17 Nov 2009, 20:29
benjiboo wrote:
So if you have 6 people going to a movie and 6 seats next to each other for them to sit, but 2 of the people, Ron and Todd, will not sit next to each other you have 6! - (5! + 5!) ways to sit everybody in this mannor. This is the total number of ways to sit 6 people, minus each of the possible ways that the two people, Ron and Todd, can sit next to each other.

First question: Is there a better strategy to solve this type of question without addition/subtraction?

Now although this may or may not be something the GMAT will ask, I am curious to the following. Lets say that of the 6 people, the two that wont sit next to each other, Ron and Todd, have decided that they WILL sit with EXACTLY 1 person in between them. How many different ways can the 6 people sit while having Ron and Todd separated by exactly 1 person? :shock:

The math/strategy behind the second part would really be appreciated.



Second Part:

Sitting arrangement
1 ---- 2 ---- 3 ---- 4 ---- 5 ---- 6
R ---- ---- T ---- ---- ---- ---- ---- = 2*4*3*2*1 = 48
---- --R ---- ---- --T ---- ---- ---- --= 2*4*3*2*1 = 48
---- ---- --- R---- ---- --T---- ---- -= 2*4*3*2*1 = 48
---- ---- ---- ---- -R---- ---- --T----= 2*4*3*2*1 = 48

So Total ways 48+48+48+48 = 192

Last edited by swatirpr on 18 Nov 2009, 05:01, edited 2 times in total.
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Re: Question on combinations and permutations [#permalink] New post 17 Nov 2009, 20:50
swatirpr wrote:
benjiboo wrote:
So if you have 6 people going to a movie and 6 seats next to each other for them to sit, but 2 of the people, Ron and Todd, will not sit next to each other you have 6! - (5! + 5!) ways to sit everybody in this mannor. This is the total number of ways to sit 6 people, minus each of the possible ways that the two people, Ron and Todd, can sit next to each other.

First question: Is there a better strategy to solve this type of question without addition/subtraction?

Now although this may or may not be something the GMAT will ask, I am curious to the following. Lets say that of the 6 people, the two that wont sit next to each other, Ron and Todd, have decided that they WILL sit with EXACTLY 1 person in between them. How many different ways can the 6 people sit while having Ron and Todd separated by exactly 1 person? :shock:

The math/strategy behind the second part would really be appreciated.



Second Part:
Seat
1 ---- 2 ---- 3 ---- 4 ---- 5 ---- 6
R ---- ---- T ---- ---- ---- ---- ---- = 2*4*3*2*1 = 48
---- --R ---- ---- --T ---- ---- ---- --= 2*4*3*2*1 = 48
---- ---- --- R---- ---- --T---- ---- -= 2*4*3*2*1 = 48
---- ---- ---- ---- -R---- ---- --T----= 2*4*3*2*1 = 48

So Total ways 48+48+48+48 = 192


Explain please. Thanks! :!:
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Re: Question on combinations and permutations [#permalink] New post 18 Nov 2009, 04:59
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benjiboo wrote:
swatirpr wrote:
benjiboo wrote:
So if you have 6 people going to a movie and 6 seats next to each other for them to sit, but 2 of the people, Ron and Todd, will not sit next to each other you have 6! - (5! + 5!) ways to sit everybody in this mannor. This is the total number of ways to sit 6 people, minus each of the possible ways that the two people, Ron and Todd, can sit next to each other.

First question: Is there a better strategy to solve this type of question without addition/subtraction?

Now although this may or may not be something the GMAT will ask, I am curious to the following. Lets say that of the 6 people, the two that wont sit next to each other, Ron and Todd, have decided that they WILL sit with EXACTLY 1 person in between them. How many different ways can the 6 people sit while having Ron and Todd separated by exactly 1 person? :shock:

The math/strategy behind the second part would really be appreciated.



Second Part:
Seat
1 ---- 2 ---- 3 ---- 4 ---- 5 ---- 6
R ---- ---- T ---- ---- ---- ---- ---- = 2*4*3*2*1 = 48
---- --R ---- ---- --T ---- ---- ---- --= 2*4*3*2*1 = 48
---- ---- --- R---- ---- --T---- ---- -= 2*4*3*2*1 = 48
---- ---- ---- ---- -R---- ---- --T----= 2*4*3*2*1 = 48

So Total ways 48+48+48+48 = 192


Explain please. Thanks! :!:



Second Part:

condition - Ron and Todd separated by exactly 1 person

So If R @ 1 then T @ 3 - 1st way
If T @ 1 then R @ 3 - 2nd way
So R n T can sit 2 ways
For remaining Seats 2, 4, 5, 6, 4 people can seat 4*3*2*1 ways
Total for this seating arrangement 2*4*3*2*1 =48

OR

If R @ 2 then T @ 4 - 1st way
If T @ 2 then R @ 4 - 2nd way
So R n T can sit 2 ways
For remaining Seats 1, 3, 5, 6, 4 people can seat 4*3*2*1 ways
Total for this seating arrangement 2*4*3*2*1 =48

OR

If R @ 3 then T @ 5 - 1st way
If T @ 3 then R @ 5 - 2nd way
So R n T can sit 2 ways
For remaining Seats 1, 2, 4, 6, 4 people can seat 4*3*2*1 ways
Total for this seating arrangement 2*4*3*2*1 =48

Sitting arrangement
1 ---- 2 ---- 3 ---- 4 ---- 5 ---- 6
R ---- ---- T ---- ---- ---- ---- ---- = 2*4*3*2*1 = 48
---- --R ---- ---- --T ---- ---- ---- --= 2*4*3*2*1 = 48
---- ---- --- R---- ---- --T---- ---- -= 2*4*3*2*1 = 48
---- ---- ---- ---- -R---- ---- --T----= 2*4*3*2*1 = 48

So Total ways 48+48+48+48 = 192

Please let me know if I am doing this wrong.
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Re: Question on combinations and permutations [#permalink] New post 21 Nov 2009, 16:37
I will get back to you. Anyone reading this post don't take the answer as correct until one of us gets back to this :)
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Re: Question on combinations and permutations [#permalink] New post 21 Nov 2009, 17:12
Expert's post
benjiboo wrote:
So if you have 6 people going to a movie and 6 seats next to each other for them to sit, but 2 of the people, Ron and Todd, will not sit next to each other you have 6! - (5! + 5!) ways to sit everybody in this mannor. This is the total number of ways to sit 6 people, minus each of the possible ways that the two people, Ron and Todd, can sit next to each other.

First question: Is there a better strategy to solve this type of question without addition/subtraction?

Now although this may or may not be something the GMAT will ask, I am curious to the following. Lets say that of the 6 people, the two that wont sit next to each other, Ron and Todd, have decided that they WILL sit with EXACTLY 1 person in between them. How many different ways can the 6 people sit while having Ron and Todd separated by exactly 1 person? :shock:

The math/strategy behind the second part would really be appreciated.


Question #1:
We have A, B, C, D, E and F. A and B don't want to sit together.

Let's count the # of ways when they sit together: glue them so that we have one unit from them {AB}. We'll have total of 5 units - {AB}{C}{D}{E}{F}. # of arrangements =5!. But we can fix {AB} as {BA} too so, 2*5!.

Total # of ways of arrangement of {A}{B}{C}{D}{E}{F}=6!.

# of arrangements when A and B will not sit together=6!-2*5!.

Question #2:
We have A, B, C, D, E and F. We want A and B to sit so that any from C, D, E and F to be between them.

Again we can fix A and B, and any X between them: so we get 4 units: {ACB}{D}{E}{F}. # of combinations 4!. {ACB} also can be {BCA}, so 2*4!. But between A and B we can place any from the four not only C so 4*2*4!.

So the final answer 4*2*4!=192
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Re: Question on combinations and permutations [#permalink] New post 21 Nov 2009, 23:52
This question cannot be done without addition/subtraction.

Reason being that there are 2 scenarios and each has their respective P&C. One scenario is for Ron/Todd to be seated at the first seat. The other scenario is when Ron/Todd are not sitting at the first seat.
Re: Question on combinations and permutations   [#permalink] 21 Nov 2009, 23:52
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