Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Hi study--At first I proceeded with treating the two letters as one unit, but then realized it doesn't work because we are only told that C must be to the right of D, not that the two must be adjacent.

With 8 letters total, without regard for uniqueness of the configurations, we have \(8!\) arrangements. Then we divide out the equivalent configurations that represent re-arranging duplicate letters A and B:

\(\frac{8!}{3!*2!}=8*7*6*5*2=3360\)

Lastly we have to apply the "C to the right of D" condition. If we have distributed the letters at random, this condition will be true half the time. So we throw out half our results to get 3360/2, or 1680.

Hi study--At first I proceeded with treating the two letters as one unit, but then realized it doesn't work because we are only told that C must be to the right of D, not that the two must be adjacent.

With 8 letters total, without regard for uniqueness of the configurations, we have \(8!\) arrangements. Then we divide out the equivalent configurations that represent re-arranging duplicate letters A and B:

\(\frac{8!}{3!*2!}=8*7*6*5*2=3360\)

Lastly we have to apply the "C to the right of D" condition. If we have distributed the letters at random, this condition will be true half the time. So we throw out half our results to get 3360/2, or 1680.

Good answer! +1. I fell into the trap of putting DC as one unit.

basically you have 8!/2!3! total no. of ways...thats 3360 now C and D have 50% probability that D is to the right of C..so 3360/2 or 1680 times is when C is to the right of D..

Hi study--At first I proceeded with treating the two letters as one unit, but then realized it doesn't work because we are only told that C must be to the right of D, not that the two must be adjacent.

With 8 letters total, without regard for uniqueness of the configurations, we have \(8!\) arrangements. Then we divide out the equivalent configurations that represent re-arranging duplicate letters A and B:

\(\frac{8!}{3!*2!}=8*7*6*5*2=3360\)

Lastly we have to apply the "C to the right of D" condition. If we have distributed the letters at random, this condition will be true half the time. So we throw out half our results to get 3360/2, or 1680.

Re: In how many different ways can the letters A, A, B, B. B, C, [#permalink]

Show Tags

30 Mar 2014, 02:52

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

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