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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.
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.
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..
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.
Re: In how many different ways can the letters A, A, B, B. B, C, [#permalink]
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]
30 Mar 2014, 02:21
Expert's post
1
This post was BOOKMARKED
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 \(\frac{3360}{2}=1680\)
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