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:

6!/(6-3)! is the formula we would use if there were no blanks or spaces.

But that just gives us the order of the people, and doesn't account for spaces between them. People could be arranged:
ABCXXX XABCXX XXABCX XXXABC
ABXCXX XABXCX XXABXC
ABXXCX XABXXC
ABXXXC

AXBCXX XAXBCX XXAXBC
AXBXCX XAXBXC
AXBXXC

AXXBCX XAXXBC
AXXBXC

AXXXBC

20 different ways

The solution is 20*6!/(6-3)!

But I don't know the formula, so I can't answer the question.

Can we think of this problem as similar to finding the possible anagrams of the word ABCXXX? A-ha!

Yes and no. If you did that, you would have all possible permutations of ABC and the three spaces, but what about C, D, and E?

Maybe do this in two steps:
1. Number of COMBINATIONS from 6c3, which is 20; multiply by
2. Number of anagrams from ABCXXX, which is 6!/(3!*1!*1!*1!), which is 120.

Total=2400.

Math is the same as my previous solution but this seems more sensible.

And the real killer is that a reasonably dedicated high school senior could this in his sleep...

Can we think of this problem as similar to finding the possible anagrams of the word ABCXXX? A-ha!

Yes and no. If you did that, you would have all possible permutations of ABC and the three spaces, but what about C, D, and E?

Maybe do this in two steps: 1. Number of COMBINATIONS from 6c3, which is 20; multiply by 2. Number of anagrams from ABCXXX, which is 6!/(3!*1!*1!*1!), which is 120.

Total=2400.

Math is the same as my previous solution but this seems more sensible.

And the real killer is that a reasonably dedicated high school senior could this in his sleep...

hey stoolfi,

we can pick 3 chairs out of 6 in 20 ways...

and then *each* of these "Ways"...seat these three..

So these three can be seated in 3! ways..isnt it.

And we have 20 of these combinations.

so...the total for 20 combination is 20 * 3! = 120 ways