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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

Question on combinations and permutations [#permalink]
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.

Re: Question on combinations and permutations [#permalink]
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?

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

Re: Question on combinations and permutations [#permalink]
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.

Re: Question on combinations and permutations [#permalink]
18 Nov 2009, 04:59

1

This post received KUDOS

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.

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

Re: Question on combinations and permutations [#permalink]
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?

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!.

Re: Question on combinations and permutations [#permalink]
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]
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.

gmatclubot

Re: Question on combinations and permutations
[#permalink]
05 Jul 2015, 19:49

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...