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 350,000 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:

Consider the expression [#permalink]
07 Sep 2010, 05:24

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

33% (02:04) correct
67% (01:51) wrong based on 9 sessions

Consider the expression \(\frac{(A!)}{((B!)^x * (C!)^y * (D!)^z))}\) where A,B,C,D,x,y,z are all positive integers >=1. Is this expression an integer ?

Re: Fun with factorials [#permalink]
08 Sep 2010, 00:27

The answer is B This is one of those questions me and a friend of mine made up to test each other (both of us have GMATs coming up). So I have a solution with me, but it is rather unconventional :

It is easy to prove that (1) alone cant be the answer since there is no constraint on x,y,z and one can make these big enough to exceed the numerator. Eg 7! / (2!^40 * 3!^20 * 1!^1)

To show that (2) alone is sufficient : Consider the question "How many permutations are possible of a set of A alphabets, of which x alphabets are each repeated B times, y alphabets each repeated C times and z alphabets each repeated D times within the set ?" The answer to this question is exactly the expression above, and we know that since it is the answer to a counting question, it must be an integer. _________________

Re: Fun with factorials [#permalink]
23 Oct 2010, 02:41

This one was so hard. Couldn't solve _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: Fun with factorials [#permalink]
25 Oct 2010, 23:43

Imagine I ask you the question : What is the number of ways you can arrange A balls, each of different color in a row ? The answer would be \(A!\)

Now I modify that question : What is the number of ways you can arrange A balls, of which B are are blue, C are red, D are green and the rest are of different but unique colors ? The answer would now be \(\frac{A!}{B!C!D!}\)

Now I modify it further : What is the number of ways you can arrange A balls, of which there are x subsets each consisting B balls each such that each subset consists of balls of a different shade of blue, and all other balls not included in these subsets are of unique colors ? The answer would now be \(\frac{A!}{B!^x}\). Notice that A has to be greater than or equal to xB

Finally I modify it a bit more : What is the number of ways you can arrange A balls, of which there are x subsets each consisting B balls each, y subsets of C balls each, z subsets of D balls each, such that each subset consists of a unique color of balls. And the rest of the set of balls are distinct from all other balls ? The answer would now be \(\frac{A!}{B!^xC!^yD!^z}\). Notice that A has to be greater than or equal to xB+yC+zD

Essentially what I am getting at is that the expression shown above is the answer to a combinatorial problem. And since the answer to a combinatorial problem is a "number of ways", such an expression always has to be an integer. _________________

safe.txmblr Can business make a difference in the great problems that we face? My own view is nuanced. I think business potentially has a significant role to play...

safe.txmblr Rebecca Henderson on the viability of the purpose-driven firm: We don’t want business people making policy...but on the other hand, what can business people do to...