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# Possible arrangements for the word REVIEW if one E can't be

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Intern
Joined: 30 Jan 2010
Posts: 13
Possible arrangements for the word REVIEW if one E can't be [#permalink]

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06 Feb 2010, 17:30
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Possible arrangements for the word REVIEW if one E can't be next to the other.
CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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06 Feb 2010, 17:51
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.

By the way, look at this problem: permutation-sitting-arrangement-90121.html
it tests the same concept.
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Intern
Joined: 30 Jan 2010
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06 Feb 2010, 18:10
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.
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06 Feb 2010, 18:14
Walker...

Tell me if I'm right

The total combination is 6! = 720
The total combination of E1 and E2 together is 5! = 120

720 - 120 = 600... BUT since they're only asking for one E, then 600/2 = 300

Am I reasoning this OK?
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06 Feb 2010, 18:34
pclg wrote:
Am I reasoning this OK?

Yeah, you are right.
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21 Feb 2010, 03:50
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.

By the way, look at this problem: permutation-sitting-arrangement-90121.html
it tests the same concept.

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!
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21 Feb 2010, 09:46
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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.

By the way, look at this problem: permutation-sitting-arrangement-90121.html
it tests the same concept.

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 $$360-120=240$$.

So yes, I think you are right.
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21 Feb 2010, 09:55
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.

By the way, look at this problem: permutation-sitting-arrangement-90121.html
it tests the same concept.

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!
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21 Feb 2010, 10:05
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.
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Joined: 22 Dec 2009
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21 Feb 2010, 10:12
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!!!
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Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

CEO
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Posts: 3584
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21 Feb 2010, 10:27
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jeeteshsingh wrote:
Hahaha! My apologises Mate!!! HIS SIDE!!!

Do you often see a woman riding a bike and developing iPhone Apps?
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21 Feb 2010, 10:36
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!!!
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JT...........
If u like my post..... payback in Kudos!!

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21 Feb 2010, 10:51

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22 Feb 2010, 08:01
pclg wrote:
Possible arrangements for the word REVIEW if one E can't be next to the other.

I just can't find the answer!

Total ways = 6!/2! = 360
When both E's together = 5! = 120

Answer = 360 - 120 = 240
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26 Oct 2010, 21:41
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

2. ExxEyy -> 4C2 * 3! * 2! = 72
3. ExxxEy -> 4C3 * 2! * 3! = 48
4. ExxxxE -> 4C4 * 4! * 1! = 24

So adding up all together -> 240
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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
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Posts: 39622
Re: Possible arrangements for the word REVIEW if one E can't be [#permalink]

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05 Nov 2013, 06:50
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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Re: Possible arrangements for the word REVIEW if one E can't be [#permalink]

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27 Nov 2015, 17:48
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Re: Possible arrangements for the word REVIEW if one E can't be   [#permalink] 27 Nov 2015, 17:48
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