combinatronics problem

First, let's ignore the restriction, as for every distinct way there is of arranging the letters so that C is to the right of D, there is exactly one way of arranging the letters so that C is to the left of D.

The number of ways of arranging n elements in a row if k are identical and the other n - k are distinct is n!/k!

For example, the letters of Madrid can be arranged in 6!/2! ways, the letters of Malaga in 6!/3! ways, and those of Zaragoza in 8!/2!3! ways.

2 A's 3 B's 1 C 1 D and 1 E can be arranged in 8!/2!3! ways = 8 x 7 x 6 x 5 x 4/2 = 56 x 60

In exactly half of these linear arrangements, 56 x 30 , the C is to the right of D

Hey Kevin,
I've come across many similar problems wherein among the letters A,A,B,B,B,C,D,E ......D,C can be replaced by X...Making the letters A,A,B,B,B,X,E and then arranged in 7!/2!3!.But I'm getting an answer of 480 so I'm missing a 4 somewhere...Also, I've obviously not arranged D,C in 2 ways coz they have to stay in that position.... I'm not able to crack this problem using that method of solving. Just hoping u show me how to do it in the above way..Thanks

Thanks for the reply Kevin.

I didn't follow the last bit of the explanation. i.e, how in exactly half of the arrangements will C be to the right of D

Thanks

i analysed the topic n found that.... - - - - - - - - suppose we try 2 make 2 letters together then there will be 7 ways to do that for example
cd------
-cd-----
--cd----
---cd---
----cd--
-----cd-
------cd
isn't my way of approach to the problem correct??? or is it wrong!!!...please answer

Ok. I see it now. Thanks Anurag!!

Anurag@Gurome wrote:

sl750 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?

The problem doesn't say that C must be positioned on the immediate right of D. C can be placed anywhere on the left of D. For example, AABCEBD or CAABBBED etc. Hence the approach of binding the letters together doesn't apply here.

The letters can be arranged among themselves in 8!/(3!*2!) = 3360 ways.
Now, in each of these arrangements C is either on the left of D or on the right. Hence, in exactly half of the arrangements C will be on the right of D.

Thus, number of arrangements such that C is on the right of D = 3360/2 = 1680

The correct answer is A.

Thanks MR.Anurag ....n i really appreciate ur help .if the question asked is to find number of arrangements such that d is on the right of c........would the answer be same as the above answer.....i need a little clarification on that