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:

Re: Premutations and Combinations [#permalink]
06 Feb 2010, 16:51

Expert's post

REVIEW

1) All arrangements: 6!/2 = 360 (1/2 in order to exclude double counting as [E1, E2] is the same as [E2,E1]) 2) All arrangements with two E together: 5!/2 = 60 3) All arrangements in which one E can't be next to the other: 360 - 60 = 300.

Re: Premutations and Combinations [#permalink]
06 Feb 2010, 17:10

Thanks walker!

I already checked the other problem, the thing is: if I use the same approach, the answer I get is 480... I really can't seem to understand WHY I should divide 6! in 2 and then 5! also.

Re: Premutations and Combinations [#permalink]
21 Feb 2010, 02:50

walker wrote:

REVIEW

1) All arrangements: 6!/2 = 360 (1/2 in order to exclude double counting as [E1, E2] is the same as [E2,E1]) 2) All arrangements with two E together: 5!/2 = 60 3) All arrangements in which one E can't be next to the other: 360 - 60 = 300.

Hi Walker / Bunuel..... I do not understand why do we divide the All E arrangements by 2

We are already considering both the E as one single component... and therefore to arrange 5 different letters - R-E1E2-V-I-W... is 5!.... I guess this number of arrangement do not include both cases like R-E1E2-V-I-W & R-E2E1-V-I-W... as we haven't multiplied 5! with 2!(ways in which E1&E2 can be arranged between themselves). Hence I don't see the need to divide 5! by 2....

Can you please let me know if my reasoning is wrong?

As per the answer should be: 6!/2! (as this arrangement as 2 duplicate E) - 5! (as duplicate E is no more concern since they are one single unit for us) = 240! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Premutations and Combinations [#permalink]
21 Feb 2010, 08:46

2

This post received KUDOS

Expert's post

jeeteshsingh wrote:

walker wrote:

REVIEW

1) All arrangements: 6!/2 = 360 (1/2 in order to exclude double counting as [E1, E2] is the same as [E2,E1]) 2) All arrangements with two E together: 5!/2 = 60 3) All arrangements in which one E can't be next to the other: 360 - 60 = 300.

Hi Walker / Bunuel..... I do not understand why do we divide the All E arrangements by 2

We are already considering both the E as one single component... and therefore to arrange 5 different letters - R-E1E2-V-I-W... is 5!.... I guess this number of arrangement do not include both cases like R-E1E2-V-I-W & R-E2E1-V-I-W... as we haven't multiplied 5! with 2!(ways in which E1&E2 can be arranged between themselves). Hence I don't see the need to divide 5! by 2....

Can you please let me know if my reasoning is wrong?

As per the answer should be: 6!/2! (as this arrangement as 2 duplicate E) - 5! (as duplicate E is no more concern since they are one single unit for us) = 240!

THEORY:

Permutations of n things of which P1 are alike of one kind, P2 are alike of second kind, P3 are alike of third kind ...................... Pr are alike of r th kind such that: P1+P2+P3+..+Pr=n is:

\(\frac{n!}{P1!*P2!*P3!*...*Pr!}\).

For example number of permutations of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutations of the letters of the word "google" is 6!/2!2!, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutations of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!.

In the original question there are 6 letters out of which E appears twice. Total number of permutations of these letters (without restriction) would be: \(\frac{6!}{2!}=360\).

# of combinations for which two E are adjacent is \(5!=120\), (consider two E as one element like: {R}{EE}{V}{I}{W}: # of permutation of these 5 elements is \(5!=120\))

Total # of permutation for which two E are not adjacent would be \(360-120=240\).

Re: Premutations and Combinations [#permalink]
21 Feb 2010, 08:55

Bunuel wrote:

jeeteshsingh wrote:

walker wrote:

REVIEW

1) All arrangements: 6!/2 = 360 (1/2 in order to exclude double counting as [E1, E2] is the same as [E2,E1]) 2) All arrangements with two E together: 5!/2 = 60 3) All arrangements in which one E can't be next to the other: 360 - 60 = 300.

Hi Walker / Bunuel..... I do not understand why do we divide the All E arrangements by 2

We are already considering both the E as one single component... and therefore to arrange 5 different letters - R-E1E2-V-I-W... is 5!.... I guess this number of arrangement do not include both cases like R-E1E2-V-I-W & R-E2E1-V-I-W... as we haven't multiplied 5! with 2!(ways in which E1&E2 can be arranged between themselves). Hence I don't see the need to divide 5! by 2....

Can you please let me know if my reasoning is wrong?

As per the answer should be: 6!/2! (as this arrangement as 2 duplicate E) - 5! (as duplicate E is no more concern since they are one single unit for us) = 240!

THEORY:

Permutations of n things of which P1 are alike of one kind, P2 are alike of second kind, P3 are alike of third kind ...................... Pr are alike of r th kind such that: P1+P2+P3+..+Pr=n is:

\(\frac{n!}{P1!*P2!*P3!*...*Pr!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is 6!/2!2!, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!.

In the original question there are 6 letters out of which E appears twice. Total number of permutation of these letters (without restriction) would be: \(\frac{6!}{2!}=360\).

# of combination for which two E are adjacent is \(5!=120\), (consider two E as one element like: {R}{EE}{V}{I}{W}: # of permutation of this 5 elements is \(5!=120\))

Total # of permutation for which two E are not adjacent would be \(360-120=240\).

So yes, I think you are right.

Thanks Bunuel.... but I doubt Walker goes wrong on Permutation & Combinations! Walker is quite good in this topic! I hope I see the same confirmation coming from her side! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Premutations and Combinations [#permalink]
21 Feb 2010, 09:05

Expert's post

jeeteshsingh wrote:

Thanks Bunuel.... but I doubt Walker goes wrong on Permutation & Combinations! Walker is quite good in this topic! I hope I see the same confirmation coming from her side!

Jeeteshsingh, from "his side" Yeah, you are absolutely right. _________________

Re: Premutations and Combinations [#permalink]
21 Feb 2010, 09:12

walker wrote:

jeeteshsingh wrote:

Thanks Bunuel.... but I doubt Walker goes wrong on Permutation & Combinations! Walker is quite good in this topic! I hope I see the same confirmation coming from her side!

Jeeteshsingh, from "his side" Yeah, you are absolutely right.

Hahaha! My apologises Mate!!! HIS SIDE!!! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Premutations and Combinations [#permalink]
21 Feb 2010, 09:36

walker wrote:

Do you often see a woman riding a bike and developing iPhone Apps?

Riding bikes is not that big a thing... but yeh... iphones apps... yeh its hard to believe...

All this while I was on the right track... unless recently I saw some of the old fourm topics and I read somewhere.. that u r a girl from Ukraine! lol! :D

No worries.. glad it's all clear now!!! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Premutations and Combinations [#permalink]
26 Oct 2010, 20:41

I solved it a bit differently;

Here my Approach goes: 1. ExEyyy Above, P(x) = 4C1 = 4. Taking ExE as 1 term along with the 3 Y's we have 4 terms that can be arranged in 4! = 24 ways. Hence the total possible combinations is: 4C1 * 4! * 1! = 96

Re: Possible arrangements for the word REVIEW if one E can't be [#permalink]
05 Nov 2013, 03:47

Hi, Just wanted to check why the answer is not 120

Review : 6 letters, 2 common = 6*5*4*3 = 360 Case where 2 Es are glued together = 5!*2 (2 as either of the two Es could come first) = 360 - 240 = 120 what am I missing

Re: Possible arrangements for the word REVIEW if one E can't be [#permalink]
05 Nov 2013, 05:50

Expert's post

adeel2000 wrote:

Hi, Just wanted to check why the answer is not 120

Review : 6 letters, 2 common = 6*5*4*3 = 360 Case where 2 Es are glued together = 5!*2 (2 as either of the two Es could come first) = 360 - 240 = 120 what am I missing

Two E's can be arranged only one way: EE. _________________

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...