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In how many different ways can the letters A, A, B, B. B, C,

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In how many different ways can the letters A, A, B, B. B, C, [#permalink] New post 15 Nov 2008, 09:02
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A
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C
D
E

Difficulty:

  55% (hard)

Question Stats:

55% (02:01) correct 45% (01:57) wrong based on 29 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-how-many-different-ways-can-the-letters-a-a-b-91460.html
[Reveal] Spoiler: OA
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Re: Combinations [#permalink] New post 15 Nov 2008, 09:45
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Hi study--At first I proceeded with treating the two letters as one unit, but then realized it doesn't work because we are only told that C must be to the right of D, not that the two must be adjacent.

With 8 letters total, without regard for uniqueness of the configurations, we have 8! arrangements. Then we divide out the equivalent configurations that represent re-arranging duplicate letters A and B:

\frac{8!}{3!*2!}=8*7*6*5*2=3360

Lastly we have to apply the "C to the right of D" condition. If we have distributed the letters at random, this condition will be true half the time. So we throw out half our results to get 3360/2, or 1680.
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Re: Combinations [#permalink] New post 15 Nov 2008, 09:51
phdizzle wrote:
Hi study--At first I proceeded with treating the two letters as one unit, but then realized it doesn't work because we are only told that C must be to the right of D, not that the two must be adjacent.

With 8 letters total, without regard for uniqueness of the configurations, we have 8! arrangements. Then we divide out the equivalent configurations that represent re-arranging duplicate letters A and B:

\frac{8!}{3!*2!}=8*7*6*5*2=3360

Lastly we have to apply the "C to the right of D" condition. If we have distributed the letters at random, this condition will be true half the time. So we throw out half our results to get 3360/2, or 1680.


Good answer! +1. I fell into the trap of putting DC as one unit.
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Re: Combinations [#permalink] New post 15 Nov 2008, 10:10
:) You get a much smaller number, as I found out the hard way, and I was shocked they didn't put it as a "trap answer"...

+1 to the OP for a good tricky question!
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Re: Combinations [#permalink] New post 15 Nov 2008, 10:45
Good solution to good tricky question i also considered DC single and fell for the trap only to find that the trap answer wasn't there. :)
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Re: Combinations [#permalink] New post 15 Nov 2008, 19:27
I get A ..

basically you have 8!/2!3! total no. of ways...thats 3360 now C and D have 50% probability that D is to the right of C..so 3360/2 or 1680 times is when C is to the right of D..

http://www.manhattangmat.com/forums/post2952.html

this link explains it well..
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Re: Combinations [#permalink] New post 15 Nov 2008, 20:25
phdizzle wrote:
Hi study--At first I proceeded with treating the two letters as one unit, but then realized it doesn't work because we are only told that C must be to the right of D, not that the two must be adjacent.

With 8 letters total, without regard for uniqueness of the configurations, we have 8! arrangements. Then we divide out the equivalent configurations that represent re-arranging duplicate letters A and B:

\frac{8!}{3!*2!}=8*7*6*5*2=3360

Lastly we have to apply the "C to the right of D" condition. If we have distributed the letters at random, this condition will be true half the time. So we throw out half our results to get 3360/2, or 1680.


Very elegant approach.....

Also a good question..

+1 for both of you..
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Re: Combinations [#permalink] New post 15 Nov 2008, 22:36
Thanks all for the explanation..
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Re: In how many different ways can the letters A, A, B, B. B, C, [#permalink] New post 30 Mar 2014, 01:52
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Re: In how many different ways can the letters A, A, B, B. B, C, [#permalink] New post 30 Mar 2014, 02:21
Expert's post
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

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: In how many different ways can the letters A, A, B, B. B, C,   [#permalink] 30 Mar 2014, 02:21
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