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# In how many ways can 5 different toys be packed in 3

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Director
Joined: 02 Mar 2006
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In how many ways can 5 different toys be packed in 3 [#permalink]  09 Nov 2006, 03:36
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In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

a.20
b.30
c.50
d.600
e.1000
Senior Manager
Joined: 01 Sep 2006
Posts: 302
Location: Phoenix, AZ, USA
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Box A B C
1 1 3 can be 3 ways= 15
1 2 2 can be 3 ways = 15

Total 30
Senior Manager
Joined: 05 Oct 2006
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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.

gmat strategy......hee hee hee...

somebody show the way plz.
SVP
Joined: 01 May 2006
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(C) for me

The 3 boxes are a kind of illusion. It's just a way to combine toys together.

So we can do 2 kinds of set:
> 2 2 1 (we separate in 3 groups : 2 toys, 2 toys and 1 toy)
> 3 1 1

(2C5)*(2C3) + (3C5)*(1C2)
= 5!/(3!*2!)*3 + 5!/(3!*2!)*2
= 5*2*3 + 5*2*2
= 50

Last edited by Fig on 09 Nov 2006, 23:16, edited 1 time in total.
Director
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Thanks fig, OA is 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).

I hope you will understand my problem!
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karlfurt wrote:
Thanks fig, OA is 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).

I hope you will understand my problem!

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.
Intern
Joined: 08 Nov 2006
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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 .
Director
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Thanks! While writing a reply because it was still not clear, I understood!

PS : I think you meant toys and not tools.

Cheers
VP
Joined: 25 Jun 2006
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andrewnorway wrote:
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 .

identical boxes!.

it is more about classification as Fig explained.
SVP
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karlfurt wrote:
Thanks! While writing a reply because it was still not clear, I understood!

PS : I think you meant toys and not tools.

Cheers

Ahhh... Late readings :p... Absolutely... They are toys not tools

Sorry on this confusing mistake.... I change all my posts now .
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