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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 .

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