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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.



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Re: Premutations and Combinations [#permalink]
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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: permutationsittingarrangement90121.htmlit tests the same concept.
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Re: Premutations and Combinations [#permalink]
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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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Re: Premutations and Combinations [#permalink]
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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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Re: Premutations and Combinations [#permalink]
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06 Feb 2010, 18:34
pclg wrote: Am I reasoning this OK? Yeah, you are right.
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Re: Premutations and Combinations [#permalink]
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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 = 603) All arrangements in which one E can't be next to the other: 360  60 = 300. By the way, look at this problem: permutationsittingarrangement90121.htmlit 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  RE1E2VIW... is 5!.... I guess this number of arrangement do not include both cases like RE1E2VIW & RE2E1VIW... 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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Re: Premutations and Combinations [#permalink]
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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 = 603) All arrangements in which one E can't be next to the other: 360  60 = 300. By the way, look at this problem: permutationsittingarrangement90121.htmlit 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  RE1E2VIW... is 5!.... I guess this number of arrangement do not include both cases like RE1E2VIW & RE2E1VIW... 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 \(360120=240\). So yes, I think you are right.
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Re: Premutations and Combinations [#permalink]
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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 = 603) All arrangements in which one E can't be next to the other: 360  60 = 300. By the way, look at this problem: permutationsittingarrangement90121.htmlit 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  RE1E2VIW... is 5!.... I guess this number of arrangement do not include both cases like RE1E2VIW & RE2E1VIW... 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 \(360120=240\). 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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Re: Premutations and Combinations [#permalink]
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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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Re: Premutations and Combinations [#permalink]
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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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Re: Premutations and Combinations [#permalink]
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21 Feb 2010, 10:27
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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Re: Premutations and Combinations [#permalink]
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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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Re: Premutations and Combinations [#permalink]
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21 Feb 2010, 10:51
Give me that link!
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Re: Premutations and Combinations [#permalink]
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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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Re: Premutations and Combinations [#permalink]
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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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Re: Possible arrangements for the word REVIEW if one E can't be [#permalink]
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