In how many different ways can the letters A, A, B : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 06:24

# LIVE NOW:

Chat with Admission Manager and Current Student of NUS SIngapore - Join Chat Room to Participate.

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In how many different ways can the letters A, A, B

Author Message
TAGS:

### Hide Tags

Manager
Joined: 02 Oct 2008
Posts: 58
Followers: 5

Kudos [?]: 60 [6] , given: 0

In how many different ways can the letters A, A, B [#permalink]

### Show Tags

24 Dec 2009, 15:17
6
KUDOS
26
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

65% (02:04) correct 35% (01:54) wrong based on 531 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2012, 01:13, edited 2 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 36554
Followers: 7078

Kudos [?]: 93172 [10] , given: 10553

### Show Tags

24 Dec 2009, 16:28
10
KUDOS
Expert's post
12
This post was
BOOKMARKED
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 $$\frac{3360}{2}=1680$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36554
Followers: 7078

Kudos [?]: 93172 [6] , given: 10553

### Show Tags

25 Dec 2009, 04:30
6
KUDOS
Expert's post
3
This post was
BOOKMARKED
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.
_________________
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 546

Kudos [?]: 3557 [2] , given: 360

### Show Tags

15 Mar 2010, 20:39
2
KUDOS
Expert's post
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 must-have app especially if you aim at 700+ | PrepGame

Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4224
Followers: 327

Kudos [?]: 3449 [1] , given: 101

### Show Tags

24 Dec 2009, 21:00
1
KUDOS
Expert's post
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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Last edited by chetan2u on 25 Dec 2009, 00:10, edited 1 time in total.
Intern
Joined: 15 Feb 2013
Posts: 5
Followers: 0

Kudos [?]: 2 [1] , given: 0

Re: In how many different ways can the letters A, A, B [#permalink]

### Show Tags

26 Jun 2013, 14:10
1
KUDOS
Possible arrangements:
1) xcxxxxxx ---- 1C1 * 6!/(2!*3!) (D can be placed in 1 position)
2) xxcxxxxx ---- 2C1 * 6!/(2!*3!) (D can be placed in 2 positions)
3) xxxcxxxx ---- 3C1 * 6!/(2!*3!) (D can be placed in 3 positions)
4) xxxxcxxx ---- 4C1 * 6!/(2!*3!) (D can be placed in 4 positions)
5) xxxxxcxx ---- 5C1 * 6!/(2!*3!) (D can be placed in 5 positions)
6) xxxxxxcx ---- 6C1 * 6!/(2!*3!) (D can be placed in 6 positions)
7) xxxxxxxc ----- 7C1 * 6!/(2!*3!) (D can be placed in 7 positions)

= (1+2+3+4+5+6+7) * 6!/(2!*3!)
= 28 * 6!/(2!*3!)
= 1680
Intern
Joined: 22 Dec 2009
Posts: 40
Followers: 0

Kudos [?]: 66 [0], given: 13

### Show Tags

24 Dec 2009, 18:48
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?
Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 770 [0], given: 33

### Show Tags

24 Dec 2009, 23:29
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 $$\frac{3360}{2}=1680$$

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
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 734 [0], given: 19

### Show Tags

25 Dec 2009, 19:30
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 $$\frac{3360}{2}=1680$$

Cannot be better than this one. +1.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 546

Kudos [?]: 3557 [0], given: 360

### Show Tags

15 Mar 2010, 17:37
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 must-have app especially if you aim at 700+ | PrepGame

Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 18

Kudos [?]: 163 [0], given: 22

### Show Tags

15 Mar 2010, 19:40
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!!!

Manager
Joined: 07 Feb 2010
Posts: 159
Followers: 2

Kudos [?]: 552 [0], given: 101

### Show Tags

06 Oct 2010, 05:33
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

Math Expert
Joined: 02 Sep 2009
Posts: 36554
Followers: 7078

Kudos [?]: 93172 [0], given: 10553

### Show Tags

06 Oct 2010, 05:48
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

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: 19 Apr 2010
Posts: 210
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 77 [0], given: 28

### Show Tags

08 Oct 2010, 03:16
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: 36554
Followers: 7078

Kudos [?]: 93172 [0], given: 10553

### Show Tags

08 Oct 2010, 03:20
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: 19 Apr 2010
Posts: 210
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 77 [0], given: 28

### Show Tags

08 Oct 2010, 03:48
Thanks Bunuel Got the point
Manager
Joined: 08 Sep 2010
Posts: 232
Location: India
WE 1: 6 Year, Telecom(GSM)
Followers: 4

Kudos [?]: 251 [0], given: 21

### Show Tags

08 Oct 2010, 06:40
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.
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1353
Followers: 17

Kudos [?]: 240 [0], given: 10

### Show Tags

30 Apr 2011, 21:07
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.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Manager
Joined: 03 Aug 2011
Posts: 241
Location: United States
Concentration: General Management, Entrepreneurship
GMAT 1: 750 Q49 V44
GPA: 3.38
WE: Engineering (Computer Software)
Followers: 1

Kudos [?]: 41 [0], given: 12

### Show Tags

17 Aug 2011, 17:13
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 re-arrange C and D, you want C to the right of D, avoiding the re-arranging 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: 149
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
Followers: 0

Kudos [?]: 15 [0], given: 103

### Show Tags

26 May 2013, 17:46
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 $$\frac{3360}{2}=1680$$

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?
Re: Permutation Problem   [#permalink] 26 May 2013, 17:46

Go to page    1   2    Next  [ 30 posts ]

Similar topics Replies Last post
Similar
Topics:
1 In how many different ways can the letters B, E, N, E, F, I, C, I, A, 1 02 Oct 2016, 04:55
2 In how many different ways can the letters of the word 6 23 Sep 2014, 03:41
15 In how many different ways can the letters of the word "CORPORATION" 6 17 Sep 2014, 11:39
7 In how many different ways can the letters A, A, B, B. B, C, 9 15 Nov 2008, 09:02
10 In how many different ways can trhee letters be posted from 11 10 Nov 2007, 12:00
Display posts from previous: Sort by