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:

Think of it like an experiment: 1. Select 3 people from 9 at random 2. Select another 3 people from the 6 remaining people 3. Select another 3 people for the 3 remaining people

The first thing to notice is that steps 1 and 2 completely determine step 3. The second thing to notice is that since we're selecting teams, order doesn't matter, so we should use combinations 9C3 = (9*8*7)/(3*2*1) =168 6C3 = (6*5*4)/(3*2*1) = 20

By the fundamental counting principle, we should multiply these two numbers to see how many outcomes our experiment has 168*20 = 3,460

Solution: The first group 9C3=84 The second group 6C3=20 And the last group 3C3=1 In total have 84∙20∙1=1680 ways, but, exist repetitions in ways, then: ( 9C3∙ 6C3∙ 3C3)/3!=1680/6=280

Second, 1/3rd of the combinations will be repetitions in 3 groups of 3 and 1/2 of the combinations will be repetitions in 2 groups of 3. So you should divided the result be 3 and 2 respectively arriving at 280.

1. Anagram: Let's focus on how to use this method, so that you can reuse it. Solution: = 9!/ (3!*3!*3!) = 1680

2. Direct and more reliable solution is to just visualize one person at a time. Solution: = (ways to select 3 out of 9) * (ways to select 3 our of remaining 6) * (ways to select 3 out of remaining 3) = 9C3*6C3*3C3 = [9!/(6!*3!)] *[6!/(3!*3!)]*1 = 9!/(3!*3!*3!) = 1680

Ans: "C"

---------------------------------------------------------- I made a mistake early on. I assumed they are 3 distinct group and hence thought order mattered. But this is NOT indicated in the question and MY ASSUMPTION WAS WRONG!

Hence the ans needs to divided by 3! (the number of ways the 3 groups can be rearranged). Hence 1680/3!=280. Ans: A

Thanks manishgeorge for pointing it out.

Hope this helps!

Last edited by abhicoolmax on 22 Jul 2011, 19:58, edited 2 times in total.