In how many different ways can the letters A, A, B : GMAT Problem Solving (PS)
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# In how many different ways can the letters A, A, B

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

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24 Dec 2009, 15:17
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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
[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
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24 Dec 2009, 16:28
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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 $$\frac{3360}{2}=1680$$

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24 Dec 2009, 18:48
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?
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24 Dec 2009, 21:00
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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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Last edited by chetan2u on 25 Dec 2009, 00:10, edited 1 time in total.
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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
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25 Dec 2009, 04:30
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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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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.
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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.
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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?
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15 Mar 2010, 20:39
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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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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

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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.
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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?
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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 $$\frac{3360}{2}=1680$$."
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08 Oct 2010, 03:48
Thanks Bunuel Got the point
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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.
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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.
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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.
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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?
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26 May 2013, 23:08
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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 $$\frac{3360}{2}=1680$$

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:
susan-john-daisy-tim-matt-and-kim-need-to-be-seated-in-130743.html
meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html
goldenrod-and-no-hope-are-in-a-horse-race-with-6-contestants-82214.html

Hope it helps.
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Re: Permutation Problem   [#permalink] 26 May 2013, 23:08

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