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**Jhua4****Member**- Registered: 2012-12-03
- Posts: 12

You have 25 cards, 15 distinguishable envelopes (i.e. envelopes are labeled 1,2,3...,15). You may put any non-negative number of cards into an envelope. In how many ways can you put the 25 cards if

a) the cards are distinguishable (e.g., if each has different message on it)

b) cards are identical

c) cards are identical and no card can be left empty

I can't figure out the answers for this problem. Help is much appreciated!:)

*Last edited by Jhua4 (2012-12-03 23:17:54)*

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**Jhua4****Member**- Registered: 2012-12-03
- Posts: 12

We have learned the following formula for putting distinguishable objects into distinguishable boxes: n!/n1!n2!...nk!

So I'm a little confused as how to apply that formula to the question.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,547

Thats for n1, n2, n3 number of cards in each each envelope..

My formula has some envelopes empty so it is not correct.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Jhua4****Member**- Registered: 2012-12-03
- Posts: 12

So what exactly is the correct answer for a)? Thanks!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,547

Hi;

Non negative means 0,1,2,3,4,5 so some boxes can be empty.

a)

ways.

b)

ways.

c)

ways.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Jhua4****Member**- Registered: 2012-12-03
- Posts: 12

Could you explain how you got to that solution please? Thanks!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,547

Hi;

Please hold on I am getting the answers to the problems. I am putting them in post #5. When I have all the computation done, I will explain the methods.

c) cards are identical and no card can be left empty

This question is incorrect.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Jhua4****Member**- Registered: 2012-12-03
- Posts: 12

Sorry I meant the following:

The cards are identical and no envelope can be left empty

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,547

Hi;

Post #5 contains all the answers you require.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Jhua4****Member**- Registered: 2012-12-03
- Posts: 12

bobbym wrote:

Hi;

Post #5 contains all the answers you require.

I see the solutions but could you also post how you got to the solution?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,547

Hi;

Basically, you are just plugging into formulas, check post #5 again.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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