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by hey_thr67 » Sun May 20, 2012 11:18 pm
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: 1,680
2: 2,160
3: 2,520
4: 3,240
5: 3,360
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by LalaB » Sun May 20, 2012 11:34 pm
8!/2!*3!*2!

u need to divide into 2!*3!*2! ,since u have 2 As , 3Bs and the letter C must be to the right of the letter D
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by Anurag@Gurome » Sun May 20, 2012 11:44 pm
hey_thr67 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: 1,680
2: 2,160
3: 2,520
4: 3,240
5: 3,360
There are 8 letters out of which A appears twice and B appears three times.
Total number of permutation of these letters (without restriction) would be 8!/(2!)(3!) = (8 * 7 * 6 * 5 * 4)/2 = 3,360.

In half of these cases D will be to the right of C and in half of these cases to the left, hence required number of ways = 3360/2 = 1,680

The correct answer is A.
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