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

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Director
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In how many ways can 5 different toys be packed in 3 [#permalink] New post 09 Nov 2006, 03:36
00:00
A
B
C
D
E

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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
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[#permalink] New post 09 Nov 2006, 07:01
Box A B C
1 1 3 can be 3 ways= 15
1 2 2 can be 3 ways = 15


Total 30
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Re: PS Permutation [#permalink] New post 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.
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 [#permalink] New post 09 Nov 2006, 14:49
(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.
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 [#permalink] New post 09 Nov 2006, 15:12
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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 [#permalink] New post 09 Nov 2006, 15:22
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!


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.
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 [#permalink] New post 09 Nov 2006, 15:35
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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 [#permalink] New post 09 Nov 2006, 15:37
Thanks! While writing a reply because it was still not clear, I understood!

PS : I think you meant toys and not tools.

Cheers
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 [#permalink] New post 09 Nov 2006, 19:31
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.
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 [#permalink] New post 09 Nov 2006, 23:15
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 :).
  [#permalink] 09 Nov 2006, 23:15
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