In how many different ways can the letters A, A, B

I don't think that by being on the right of D they mean C has to be right next to it. C can be anywhere. Therefore you can't "tie" them together and make it into 7 slots. You have to consider all 8 slots.
Hope I helped!

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

Hi..
First let's calculate total ways..
What do we have ..
2*A, 3*B and 1 of C,D and E..
So total 8..
Ways of combination= 8!/2!3!=8*7*5*4*3=3360..
But in these half will have C on right of D and half D on right of C..
So ans = 3360/2=1680

A
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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)

Can someone help me with this problem . really not able to understand the logic?

Check the previous two pages. For example:
https://gmatclub.com/forum/in-how-many- ... ml#p666849
https://gmatclub.com/forum/in-how-many-d ... ml#p666937
https://gmatclub.com/forum/in-how-many-d ... ml#p794968
https://gmatclub.com/forum/in-how-many-d ... ml#p796306

Similar questions:
https://gmatclub.com/forum/susan-john-da ... 30743.html
https://gmatclub.com/forum/meg-and-bob-a ... 58095.html
https://gmatclub.com/forum/six-mobsters- ... 26151.html
https://gmatclub.com/forum/mary-and-joe- ... 26407.html
https://gmatclub.com/forum/goldenrod-and ... 82214.html

Hope it helps.
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The question says C has to be to the right of D not necessarily in consequtive places. So if D takes the first position then C has 7 positions as there are 8 places. Then if D takes 2nd place there are 6 places to it's right for C. If D is in 3rd position than C has 5 and so on and the last position D can take is the 7th place so that C can take the 8th place. So in all D with C to it's right is possible in 7+6+5+4+3+2+1=28 positions. For each of these, the rest 6 can arrange themselves in 6!. Since there are repetitions, the total ways for the rest 6 of them will be 6!/(2!*3!)
So the total arrangements with C to the right of D will be 28*6!/(2!*3!)= 1680. So A

Posted from my mobile device

After reviewing this problem and this similar one that Bunuel suggested (https://gmatclub.com/forum/susan-john-d ... 30743.html) I finally understand the hidden trap in the wording of this question.

The way this question is worded made me assume that C had to be right next to D, which made the problem much more confusing than it actually is. In fact the wording is just that C has to be TO THE RIGHT OF D, which means C and D could be respectively in any position in the row as long as C was right of D's position.

Therefore just need to calculate total # of ways A, A, B, B, B, C, D, E can be arranged, and then divide by 2 since the probability of C being to the right of D occurs half of the time compared to D being to the right of C.

\(\frac{8!}{(2! * 3!)} = \frac{3360}{2} = 1680\)

Hi everyone, I formula it differently and would like to know why it has to multiple 2

8 letters
so
7! x 2!/2!3! = 840
(since C & D counts as one unit)
but the answer is 840 x2, why does it have to multiple 2?

TIA!

Total number of unique arrangements = 8! /(2!3!) = 3360.

Half of the time C will be to the right of D. 3360/2 = 1680.

Correct answer is A.

Can somebody please point out where I went wrong with my following approach?
Leave alone C and D for a moment, we have 6 alphabets with certain repetitions. Total number of their arrangements possible:
6!/(2!*3!) = 60
Now, any arrangement of 6 characters leaves 7 spaces. No. of ways to choose 2 spaces out of those 7:
7C2 = 21
Total arrangements, thus: 60 * 21 = 1260
....
Also, in one of the solutions provided above it has been mentioned that of all the arrangements possible without any restriction, half of arrangements will have C on left of D and the other half other way. This seems logical by I'm not able to comprehend or visualize this for some reason. Any help will be appreciated.

Regards
Apourv

Thank you for the detailed explanation Bunuel

Bunuel !

i tried solving this question this way please tell me where i went wrong:
A A B B B (DC) E i have consider DC as one quantity should be only this way 1!
now 7!/ 2! * 3! Bunuel

Bunuel
AABBB (DC) E ,
so can we solve this way 7!/3! * 2! , where is it going wrong , i understand your solution but still !!

The concept of Symmetry in Arrangements:

No. of Total Arrangements in which D appears before C = No. of arrangements in which D appears AFTER C


In other word, C will be to the RIGHT of D in (1/2) of the Total Possible Arrangements

In the other (1/2) of Total Possible Arrangements, C will appear to the LEFT of D


total no of arrangements of 8 elements, of which 2 A Elements are Identical and 3 B Elements are Identical =

8! / (2! * 3!) = 4 * 7 * 6 * 5 * 4

(1/2) of these total arrangements will give us the arrangements we want:

(1/2) * (4 * 7 * 6 * 5 * 4) = 1,680

Posted from my mobile device

Hi Bunuel. How would you solve in the situation where there are 2Ds and 1 or 2 Cs, and the question remains the same -- how many possibilities where Cs are to the right of Ds? (Basically the same scenario but extended for multiple letters?) Or even if given a combo of three letters having a specific relationship

This was my logic too. Can anyone explain why this isn't the correct approach?

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