May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 02 Oct 2008
Posts: 49

In how many different ways can the letters A, A, B
[#permalink]
Show Tags
Updated on: 17 Jun 2012, 02:13
Question Stats:
64% (01:42) correct 36% (02:19) wrong based on 610 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by tania on 24 Dec 2009, 16:17.
Last edited by Bunuel on 17 Jun 2012, 02:13, edited 2 times in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 55266

Re: Permutation Problem
[#permalink]
Show Tags
24 Dec 2009, 17:28
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.
_________________




Intern
Joined: 22 Dec 2009
Posts: 38

Re: Permutation Problem
[#permalink]
Show Tags
24 Dec 2009, 19:48
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?



Math Expert
Joined: 02 Aug 2009
Posts: 7684

Re: Permutation Problem
[#permalink]
Show Tags
Updated on: 25 Dec 2009, 01:10
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
_________________
Originally posted by chetan2u on 24 Dec 2009, 22:00.
Last edited by chetan2u on 25 Dec 2009, 01:10, edited 1 time in total.



Director
Joined: 03 Sep 2006
Posts: 769

Re: Permutation Problem
[#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 55266

Re: Permutation Problem
[#permalink]
Show Tags
25 Dec 2009, 05:30
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.
_________________



SVP
Joined: 29 Aug 2007
Posts: 2312

Re: Permutation Problem
[#permalink]
Show Tags
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.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



CEO
Joined: 17 Nov 2007
Posts: 3408
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: Probability Q
[#permalink]
Show Tags
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.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Senior Manager
Joined: 21 Jul 2009
Posts: 313
Schools: LBS, INSEAD, IMD, ISB  Anything with just 1 yr program.

Re: Probability Q
[#permalink]
Show Tags
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?
_________________
I am AWESOME and it's gonna be LEGENDARY!!!



CEO
Joined: 17 Nov 2007
Posts: 3408
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: Probability Q
[#permalink]
Show Tags
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.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Manager
Joined: 07 Feb 2010
Posts: 127

Re: Probability Q
[#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 55266

Re: Probability Q
[#permalink]
Show Tags
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.
_________________



Manager
Joined: 20 Apr 2010
Posts: 174
Schools: ISB, HEC, Said

Re: Probability Q
[#permalink]
Show Tags
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?



Math Expert
Joined: 02 Sep 2009
Posts: 55266

Re: Probability Q
[#permalink]
Show Tags
08 Oct 2010, 04:20
prashantbacchewar wrote: Bunuel.. I am confused here.
How do you say that in all the 3360 cases C and D will be together? I'm not saying that anywhere. "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 \(\frac{3360}{2}=1680\)."
_________________



Manager
Joined: 20 Apr 2010
Posts: 174
Schools: ISB, HEC, Said

Re: Probability Q
[#permalink]
Show Tags
08 Oct 2010, 04:48
Thanks Bunuel Got the point



Manager
Joined: 08 Sep 2010
Posts: 165
Location: India
WE 1: 6 Year, Telecom(GSM)

Re: Probability Q
[#permalink]
Show Tags
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.



Director
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 966

Re: Probability Q
[#permalink]
Show Tags
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.



Manager
Joined: 03 Aug 2011
Posts: 213
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.38
WE: Engineering (Computer Software)

Re: Probability Q
[#permalink]
Show Tags
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.



Manager
Joined: 14 Nov 2011
Posts: 117
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

Re: Permutation Problem
[#permalink]
Show Tags
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?



Math Expert
Joined: 02 Sep 2009
Posts: 55266

Re: Permutation Problem
[#permalink]
Show Tags
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.
_________________




Re: Permutation Problem
[#permalink]
27 May 2013, 00:08



Go to page
1 2
Next
[ 28 posts ]



