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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.

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.

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|

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 \(360-120=240\).

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|

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.
_________________

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|

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|

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]

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05 Nov 2013, 04:47

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

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.
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Taking different conditions where one letter is between the two "E"s, then a condition where two letters are between the two "E"s, then 3 and then finally 4 between the two "E". Total combinations= 4C1*4!+ 4C2*2*3!+4C3*3!*2+4C4*4! = 96+72+48+24 = 240.