It is currently 26 Jun 2017, 18:51

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Question on combinations and permutations

Author Message
TAGS:

### Hide Tags

Intern
Joined: 24 Sep 2009
Posts: 23
Question on combinations and permutations [#permalink]

### Show Tags

11 Nov 2009, 10:27
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.
Manager
Joined: 13 Oct 2009
Posts: 118
Location: USA
Schools: IU KSB
Re: Question on combinations and permutations [#permalink]

### Show Tags

17 Nov 2009, 21:29
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.

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, 06:01, edited 2 times in total.
Intern
Joined: 24 Sep 2009
Posts: 23
Re: Question on combinations and permutations [#permalink]

### Show Tags

17 Nov 2009, 21:50
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

Manager
Joined: 13 Oct 2009
Posts: 118
Location: USA
Schools: IU KSB
Re: Question on combinations and permutations [#permalink]

### Show Tags

18 Nov 2009, 05:59
1
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.

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

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.
Intern
Joined: 24 Sep 2009
Posts: 23
Re: Question on combinations and permutations [#permalink]

### Show Tags

21 Nov 2009, 17:37
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
Math Expert
Joined: 02 Sep 2009
Posts: 39702
Re: Question on combinations and permutations [#permalink]

### Show Tags

21 Nov 2009, 18:12
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!.

_________________
Intern
Joined: 22 Nov 2009
Posts: 1
Re: Question on combinations and permutations [#permalink]

### Show Tags

22 Nov 2009, 00:52
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.
Intern
Joined: 21 Jun 2014
Posts: 3
Re: Question on combinations and permutations [#permalink]

### Show Tags

05 Jul 2015, 20:49
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.
Re: Question on combinations and permutations   [#permalink] 05 Jul 2015, 20:49
Similar topics Replies Last post
Similar
Topics:
Permutation and Combination Basic Question ??????? 3 21 Jul 2016, 12:59
1 Permutation and Combination Conceptual Question 4 20 May 2014, 07:38
permutation and combination questions 2 04 Aug 2013, 00:57
Skipping Permutation/Combination questions 1 10 Jan 2010, 13:03
99 Permutations, Combinations, Probability - Download Questions 83 06 Jul 2016, 04:23
Display posts from previous: Sort by