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benjiboo
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Question on combinations and permutations 11 Nov 2009, 10:27
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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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Question on combinations and permutations Updated on: 18 Nov 2009, 06:01
Originally posted by swatirpr on 17 Nov 2009, 21:29.
Last edited by swatirpr on 18 Nov 2009, 06:01, edited 2 times in total.
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benjiboo
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.
Show more


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
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Question on combinations and permutations 17 Nov 2009, 21:50
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swatirpr
benjiboo
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
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Explain please. Thanks! :!:
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Question on combinations and permutations 18 Nov 2009, 05:59
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benjiboo
swatirpr
benjiboo
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! :!:
Show more


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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Question on combinations and permutations 21 Nov 2009, 17:37
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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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Question on combinations and permutations 21 Nov 2009, 18:12
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benjiboo
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.
Show more

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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Question on combinations and permutations 22 Nov 2009, 00:52
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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.
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Question on combinations and permutations 05 Jul 2015, 20:49
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benjiboo
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.
Show more

Ans for First Que.: It is itself a better strategy......there is no need to go for alternate when time matters.

Ans for Second Que. (Part) :

Consider that Ron, Todd and one of their friend(say X) have a single big seat and now the total available seats are 04.

Now in this case total combination will be 4! where it doesn’t matter Ron, Todd and X occupy which sitting order on the single big seat (viz. X is in between Ron and Todd or not).

Now, in case of X sitting in between Ron and Todd, no. of combination = 2 * 4!

Since, this order can be made with total four friends (including X) the final combination will be= 4 * 2* 4! = 192.
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Question on combinations and permutations 13 Jul 2018, 06:34
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First part of the question:

so the strategy is to compute the total ways of seating and subtract the number of ways they sit together.

total ways=6! ... this i understand...

# of ways they sit together:

a,b,c,d,[rt]

now, how many ways can they sit together?

1. [rt],a,b,c,d
2. a,[rt],b,c,d
3. a,b,[rt],c,d
4. a,b,c,[rt],d
5. a,b,c,d,[rt]

that is 5 ways not 5!.....so rt can be swapped which makes 2 x 5 = 10 ways.

can someone explain what is wrong with my work above?

thanks
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Question on combinations and permutations 20 Aug 2020, 23:51
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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Thank you for understanding, and happy exploring!