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Question on combinations and permutations
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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? The math/strategy behind the second part would really be appreciated.



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Re: Question on combinations and permutations
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Updated on: 18 Nov 2009, 05:01
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? 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
Originally posted by swatirpr on 17 Nov 2009, 20:29.
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
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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? 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
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18 Nov 2009, 04:59
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? 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
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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
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21 Nov 2009, 17:12
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? 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
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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.



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Re: Question on combinations and permutations
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05 Jul 2015, 19:49
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? The math/strategy behind the second part would really be appreciated. 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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Re: Question on combinations and permutations
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13 Jul 2018, 05:34
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




Re: Question on combinations and permutations &nbs
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13 Jul 2018, 05:34






