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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? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

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 \(\frac{3360}{2}=1680\)

why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?

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
_________________

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? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

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.

Can you please explain the logic or how you could deduce quickly the following:-

Now, in half of these cases D will be to the right of C and in half of these cases to the left

why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?

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: \(\frac{8!}{2!3!}=3360\).

LM wrote:

Can you please explain the logic or how you could deduce quickly the following:-

Now, in half of these cases D will be to the right of C and in half of these cases to the left

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.

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?

A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

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.

Cannot be better than this one. +1.
_________________

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.
_________________

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.

Can you be more clear in your explanation with the step 3?

I considered C to the right of D, the combination together as one unit and there are 7 units to be arranged with 2 A's and 3 B's. Obviously, I was not even close to any of the options. What's wrong with my approach?
_________________

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: .

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

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: .

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

CAN ANYONE EXPLAIN LAST STEP

THANKS IN ADVANCE

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.

How do you say that in all the 3360 cases C and D will be together?

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 \(\frac{3360}{2}=1680\)."
_________________

How do you say that in all the 3360 cases C and D will be together?

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.

Out of 8 letters, 2 A's and 3 B's can be arranged in 2! and 3! ways respectively. Hence total ways = 8! / 2! * 3! Now in half of them C will be towards the left of D. So for eliminating that we divide by 2. 8! / 2! * 3! *2 = 1680. Hence A.
_________________

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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.

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? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

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.

Hi Bunnel,

If the question were:

Cases in which A is to the right of C? tot=8!/2!*3!=3360

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? A.1680 B.2160 C.2520 D.3240 E.3360

Can someone explain how I should approach to solve the above problem?

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.

Hi Bunnel,

If the question were:

Cases in which A is to the right of C? tot=8!/2!*3!=3360

A to right of C = 3360/(2!)^2?

Both A's or just one A? Anyway, in this case the problem will be out of the scope of the GMAT, so I wouldn't worry about it.

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