We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \frac{8!}{2!3!}=3360.

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \frac{3360}{2}=1680

Answer: A.
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there are 8 nos in total therefore there are 8! ways to arrange them....
however A is used two times and B three times.....so A can be arranged within itself 2! and similarly B 3! times.
since all B and A are the same the times these can be arranged within themselves are the same combination ...
so distinct combinations would be 8!/2!3!...
and ofcourse the restr is c on right of d... it is clear that in half cases it is possible so divide the soln above by 2

Permutations of n things of which P1 are alike of one kind, P2 are alike of second kind, P3 are alike of third kind ...................... Pr are alike of r th kind such that: P1+P2+P3+..+Pr=n is:

\frac{n!}{P1!*P2!*P3!*...*Pr!}.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is 6!/2!2!, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!.

In the original question there are 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \frac{8!}{2!3!}=3360.



Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else?

Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left.

Hope it's clear.
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A

1) the total number of arrangements: 8!
2) excluding double counting (A1, A2 and A2, A1 are the same): 8!/2!*3! = 3360
3) the number of arrangements with C D is equal the number of arrangements with D C. Therefore, answer is 3360/2 = 1680.
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3) We always have twins, for example ???C?D?? - ???D?C??. In other words, exactly half of all arrangements has C to the right of the letter D.

Your approach considers only arrangements in which D and C are together. But we can have letters between D and C.
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Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else?

Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left.

Hope it's clear.
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I'm not saying that anywhere.

"We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \frac{8!}{2!3!}=3360.

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \frac{3360}{2}=1680."
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In your approach you are thinking that D and C should be together But C can be anywhere after D.I mean If D is at the first position C can be at any position 2nd,3rd,4th or so on.

Consider KUDOS if u like the explanation.

(1/2)(8!/(2! * 3!)) as in other half we have C to the left of D.

=1680

Answer is A.

another way to think of it is you have 8 spaces, and you want to place C and D somewhere in those 8 spaces.

if you do 8c2 * 2! for example, you're saying there are 8 spots, i want to choose two of them for C and D. Also, i want to multiple by 2! because that is # of ways I can arrange C and D, ie C D, D C. But then you realize, you don't want to re-arrange C and D, you want C to the right of D, avoiding the re-arranging means just getting rid of the 2!, same as dividing by 2

(8c2 *2!) / 2

these are all the cases that C is to the right of D, and as Bunuel pointed out, it's also the same # of cases as D to the right of C.

You can just get the total cases now and subtract the above from it.