Possible arrangements:
1) xcxxxxxx ---- 1C1 * 6!/(2!*3!) (D can be placed in 1 position)
2) xxcxxxxx ---- 2C1 * 6!/(2!*3!) (D can be placed in 2 positions)
3) xxxcxxxx ---- 3C1 * 6!/(2!*3!) (D can be placed in 3 positions)
4) xxxxcxxx ---- 4C1 * 6!/(2!*3!) (D can be placed in 4 positions)
5) xxxxxcxx ---- 5C1 * 6!/(2!*3!) (D can be placed in 5 positions)
6) xxxxxxcx ---- 6C1 * 6!/(2!*3!) (D can be placed in 6 positions)
7) xxxxxxxc ----- 7C1 * 6!/(2!*3!) (D can be placed in 7 positions)

Adding them all
= (1+2+3+4+5+6+7) * 6!/(2!*3!)
= 28 * 6!/(2!*3!)
= 1680

I didn't understand the statement "Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left." Asked myself Why? How? (similar situation with the arrangement of Frankie and Joe - question)
The following helped me understand: The letters A, B and C can be arranged in 3 slots in 3*2*1 ways.
ABC, ACB, BAC, BCA, CAB and CBA. Under normal circumstances C comes before B thrice; B comes before C thrice. For a specific case we consider whatever is required.

I hope my understanding is right. If yes, I hope it helps someone.

-Arvind

These are the arrangements possible where letter C is to the right of D
D******C
D*****C*
D****C**
D***C***
D**C****
D*C*****
DC******(7 combinations when D is the first alphabet)
*D*****C
*D****C*
*D***C**
*D**C***
*D*C****
*DC*****(6 combinations when D is the second alphabet)
**D****C
**D***C*
**D**C**
**D*C***
**DC****(5 combinations when D is the third alphabet)
The total arrangements possible are \(7+6+5+4+3+2+1 = 28\)

The total ways in which alphabets A,A,B,B,B,C can be arranged are\(\frac{6!}{2!*3!} = \frac{6*5*4*3*2}{2*3*2} = 6*5*2 = 60\)

Therefore, the total ways in which the alphabets can be arranged when C is to the right of D is 28*60 = 1680(Option A)
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Check the previous two pages. For example:
https://gmatclub.com/forum/in-how-many- ... ml#p666849
http://gmatclub.com/forum/in-how-many-d ... ml#p666937
http://gmatclub.com/forum/in-how-many-d ... ml#p794968
http://gmatclub.com/forum/in-how-many-d ... ml#p796306

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Hope it helps.
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