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

Question

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

phdizzle · Answer

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.

icandy · Answer

Good answer! +1. I fell into the trap of putting DC as one unit.

phdizzle · Answer

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!

hibloom · Answer

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.

FN · Answer

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..

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

this link explains it well..

GMAT TIGER · Answer

Very elegant approach.....

Also a good question..

+1 for both of you..

study · Answer

Thanks all for the explanation..

Bunuel · Answer

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

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