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:

Re: PS Permutation [#permalink]
09 Nov 2006, 10:23

each of the toy can goto any of the 3 boxes a,b or c...in 3 ways..
so total no of ways in which 5 toys can be assigned to 3 boxes is

3*3*3..5 times = 243...now this includes cases these toys might not goto one or 2 boxes..
so i guess the ans should be not d or e..!!!
i wud go with choice c.

But can you please explain the second situation 3 1 1

That means that we calculate the number of ways to take 3 toys from 5 and put them in the box. So 2 toys and 2 empty identical boxes are remaining.
That's what disturbs me : you say there are two ways to put these 2 toys in the two boxes, or 1C2. But if the boxes are the same, let say A and A', it doesn't matter that one of the toy (X) goes in A or A'. I better explain with following:

X is in A and Y is in A' OR X is in A' and Y is in A. since A=A', there is finally only one possibility:

Like : How many words can I write with the letters A and A: only one! and not two(AA' or A'A).

But can you please explain the second situation 3 1 1

That means that we calculate the number of ways to take 3 toys from 5 and put them in the box. So 2 toys and 2 empty identical boxes are remaining. That's what disturbs me : you say there are two ways to put these 2 toys in the two boxes, or 1C2. But if the boxes are the same, let say A and A', it doesn't matter that one of the toy (X) goes in A or A'. I better explain with following:

X is in A and Y is in A' OR X is in A' and Y is in A. since A=A', there is finally only one possibility:

Like : How many words can I write with the letters A and A: only one! and not two(AA' or A'A).

I hope you will understand my problem!

I understand your question

Well, I think the bold is not correct... The fact that A=A' in their propertities but does not mean we have only 1 possibility ; A and A' are 2 toys .... We need to select one to put it in the second box

It's where boxes are important (and no more an illusion ) .... We have to select 1 toy among 2 remaing, identical toys (combination) to put in the second box

Last edited by Fig on 09 Nov 2006, 23:16, edited 1 time in total.

I am on a different track here. The boxes are distinct, so you can't just have 221 and 311, you would also need 212, 113, etc. If you were just figuring out the different ways you could arrange 5 toys into 3 boxes, you would have 60 different ways. In each of those ways you have 2 toys left over to spread around the boxes, so I think you need to multiply 60 X the number of ways you can distribute 2 toys between the 3 boxes. Which is giving me 60X9 and not one of the answers, so I am missing something. But if it was the gmat, I would just take the answer 600 and move on .

I am on a different track here. The boxes are distinct, so you can't just have 221 and 311, you would also need 212, 113, etc. If you were just figuring out the different ways you could arrange 5 toys into 3 boxes, you would have 60 different ways. In each of those ways you have 2 toys left over to spread around the boxes, so I think you need to multiply 60 X the number of ways you can distribute 2 toys between the 3 boxes. Which is giving me 60X9 and not one of the answers, so I am missing something. But if it was the gmat, I would just take the answer 600 and move on .

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...