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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

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

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

Hi;

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

a)

ways.

b)

ways.

c)

ways.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

Hi;

Post #5 contains all the answers you require.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

Hi;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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