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In how many different ways can the letters A, A, B [#permalink]
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In how many different ways can the letters A, A, B, B, B, C, D, E be arranged if the letter C must be to the right of the letter D? A. 1680 B. 2160 C. 2520 D. 3240 E. 3360
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Last edited by Bunuel on 17 Jun 2012, 02:13, edited 2 times in total.
Edited the question and added the OA



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Re: Permutation Problem [#permalink]
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tania wrote: In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360
Can someone explain how I should approach to solve the above problem? We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\). Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\) Answer: A.
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Re: Permutation Problem [#permalink]
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24 Dec 2009, 19:48
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?



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Re: Permutation Problem [#permalink]
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24 Dec 2009, 22:00
gmatJP wrote: why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!? there are 8 nos in total therefore there are 8! ways to arrange them.... however A is used two times and B three times.....so A can be arranged within itself 2! and similarly B 3! times. since all B and A are the same the times these can be arranged within themselves are the same combination ... so distinct combinations would be 8!/2!3!... and ofcourse the restr is c on right of d... it is clear that in half cases it is possible so divide the soln above by 2
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Re: Permutation Problem [#permalink]
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25 Dec 2009, 00:29
Bunuel wrote: tania wrote: In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360
Can someone explain how I should approach to solve the above problem? We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\). Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\) Answer: A. Can you please explain the logic or how you could deduce quickly the following: Now, in half of these cases D will be to the right of C and in half of these cases to the left



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gmatJP wrote: why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!? 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 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\). LM wrote: Can you please explain the logic or how you could deduce quickly the following:
Now, in half of these cases D will be to the right of C and in half of these cases to the left Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else? Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left. Hope it's clear.
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Re: Permutation Problem [#permalink]
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25 Dec 2009, 20:30
Bunuel wrote: tania wrote: In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D?
A.1680 B.2160 C.2520 D.3240 E.3360
Can someone explain how I should approach to solve the above problem? We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\). Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\) Answer: A. Cannot be better than this one. +1.
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Re: Probability Q [#permalink]
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15 Mar 2010, 18:37
A 1) the total number of arrangements: 8! 2) excluding double counting (A1, A2 and A2, A1 are the same): 8!/2!*3! = 3360 3) the number of arrangements with C D is equal the number of arrangements with D C. Therefore, answer is 3360/2 = 1680.
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Re: Probability Q [#permalink]
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15 Mar 2010, 20:40
walker wrote: A
1) the total number of arrangements: 8! 2) excluding double counting (A1, A2 and A2, A1 are the same): 8!/2!*3! = 3360 3) the number of arrangements with C D is equal the number of arrangements with D C. Therefore, answer is 3360/2 = 1680. Can you be more clear in your explanation with the step 3? I considered C to the right of D, the combination together as one unit and there are 7 units to be arranged with 2 A's and 3 B's. Obviously, I was not even close to any of the options. What's wrong with my approach?
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Re: Probability Q [#permalink]
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15 Mar 2010, 21:39
3) We always have twins, for example ???C?D??  ???D?C??. In other words, exactly half of all arrangements has C to the right of the letter D. Your approach considers only arrangements in which D and C are together. But we can have letters between D and C.
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Re: Probability Q [#permalink]
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06 Oct 2010, 06:33
We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: .
Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be
CAN ANYONE EXPLAIN LAST STEP
THANKS IN ADVANCE



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Re: Probability Q [#permalink]
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06 Oct 2010, 06:48
anilnandyala wrote: We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: .
Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be
CAN ANYONE EXPLAIN LAST STEP
THANKS IN ADVANCE Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else? Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left. Hope it's clear.
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Re: Probability Q [#permalink]
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08 Oct 2010, 04:16
Bunuel.. I am confused here.
How do you say that in all the 3360 cases C and D will be together?



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08 Oct 2010, 04:20



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08 Oct 2010, 04:48
Thanks Bunuel Got the point



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Re: Probability Q [#permalink]
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08 Oct 2010, 07:40
prashantbacchewar wrote: Bunuel.. I am confused here.
How do you say that in all the 3360 cases C and D will be together? In your approach you are thinking that D and C should be together But C can be anywhere after D.I mean If D is at the first position C can be at any position 2nd,3rd,4th or so on. Consider KUDOS if u like the explanation.



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Re: Probability Q [#permalink]
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30 Apr 2011, 22:07
Out of 8 letters, 2 A's and 3 B's can be arranged in 2! and 3! ways respectively. Hence total ways = 8! / 2! * 3! Now in half of them C will be towards the left of D. So for eliminating that we divide by 2. 8! / 2! * 3! *2 = 1680. Hence A.
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Re: Probability Q [#permalink]
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17 Aug 2011, 18:13
another way to think of it is you have 8 spaces, and you want to place C and D somewhere in those 8 spaces.
if you do 8c2 * 2! for example, you're saying there are 8 spots, i want to choose two of them for C and D. Also, i want to multiple by 2! because that is # of ways I can arrange C and D, ie C D, D C. But then you realize, you don't want to rearrange C and D, you want C to the right of D, avoiding the rearranging means just getting rid of the 2!, same as dividing by 2
(8c2 *2!) / 2
these are all the cases that C is to the right of D, and as Bunuel pointed out, it's also the same # of cases as D to the right of C.
You can just get the total cases now and subtract the above from it.



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Re: Permutation Problem [#permalink]
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26 May 2013, 18:46
Bunuel wrote: tania wrote: In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360
Can someone explain how I should approach to solve the above problem? We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\). Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\) Answer: A. Hi Bunnel, If the question were: Cases in which A is to the right of C? tot=8!/2!*3!=3360 A to right of C = 3360/(2!)^2?



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Re: Permutation Problem [#permalink]
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27 May 2013, 00:08
cumulonimbus wrote: Bunuel wrote: tania wrote: In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D? A.1680 B.2160 C.2520 D.3240 E.3360
Can someone explain how I should approach to solve the above problem? We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\). Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\) Answer: A. Hi Bunnel, If the question were: Cases in which A is to the right of C? tot=8!/2!*3!=3360 A to right of C = 3360/(2!)^2? Both A's or just one A? Anyway, in this case the problem will be out of the scope of the GMAT, so I wouldn't worry about it. Questions about the same concept to practice: susanjohndaisytimmattandkimneedtobeseatedin130743.htmlmegandbobareamongthe5participantsinacyclingrace58095.htmlsixmobstershavearrivedatthetheaterforthepremiereofthe126151.htmlmaryandjoearetothrowthreediceeachthescoreisthe126407.htmlgoldenrodandnohopeareinahorseracewith6contestants82214.htmlHope it helps.
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