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by cans » Tue Sep 20, 2011 12:55 am
IMO A
AABBBCDE
total arrangements: 8!/3!2!
Thus C right to D: 8!/3!2!2 (in half it will be to the left of D and in other half to the right of D)
thus 1680
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by Anurag@Gurome » Tue Sep 20, 2011 2:18 am
jogi1984 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?


1,680
2,160
2,520
3,240
3,360
There are 8 letters out of which A appears 2 times and B appears 3 times.
So, without any restriction, total arrangements = 8!/(2! * 3!) = 3,360
In half of the cases D will be to the right of C and in half of the cases it will to the left of C, so the required no. of ways = 3,360/2 = 1,680

The correct answer is A.
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