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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

What are the chances that 6 people celebrate their Birthday in the same 2 months? Assume all months are equal.

*Last edited by anna_gg (2013-03-24 01:27:25)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi;

*Last edited by bobbym (2013-03-24 01:13:40)*

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

bobbym wrote:

Hi;

See my corrected description.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi;

That is a little different. I hope I am understanding what you want. It seems you want all six people to have their birthdays in only two months.

For that I am getting.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Right

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi;

Whamo! Wunderbar! Doron Zeilberger eat your heart out. Am I the king of comby/proby or what?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

How'd you get that answer?

And, I am pretty sure his last name is Zielberger.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi;

No, it is Zeilberger. The way I get all the answers I get.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

Yes. I permuted the i and the e. It seems to be common with me. I have a tough time remembering whether it's Liebniz or Leibniz.

Direct count?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

No, all combinatoric problems fall into categories. Surely you have read the 12 fold way or even better the 30 fold way. I have my own set in addition to those. These templates help solve many types of problems.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

Would you mind sharing?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Of course you were half right I already had the answer before I even began.

Of course, I will put down the formula-template here.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

Great, thank you!

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

This looks like some of Feller's work or maybe Rose, I am not sure.

Surely you recognize that?!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

What the hell is that?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Looks like a formula! So you do not recognize it?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

It looks like some stuff I've seen on Wiki's distributions pages.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Better than that. With that you compute the answer to anna's problem quickly.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

I will try to share my solution in a more explanatory way (Sorry, Bobby, I don't imply that yours is not easy to understand - it is just that I am a novice and not very familiar with complicated solutions!!):

We first must calculate all the possible ways to get 2 months out of 12, that is

Then we must calculate all the different ways by which we can arrange the birthdays of 6 people in these 2 months: Either 5 people have their birthday in the first month and 1 in the second, or 4 in the first and 2 in the second etc. Obviously we do not consider the case of 0/6 or 6/0. For the first case, we first get 1 out of 6 (for the first persons birthday) and then for the second persons it will be 5 out of 5, and so on.

Here is the calculation:

The total probability is the product of the first two (66 x 62) divided by the total number of all different ways by which 6 people can have their birthdays in 12 different months, that is, 12^6.

So we have

*Last edited by anna_gg (2013-03-30 04:58:57)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi anna_gg;

Sorry, Bobby, I don't imply that yours is not easy to understand

No problem. I am glad to see your solution. Also, anyone who can do that problem I do not characterize as a novice.

Have you tried Codecogs?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

No I haven't; actually I wrote the solution in a Word doc, but when I copied it here, the formatting was screwed up (

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi anna_gg;

I latexed it up for you.

When you have time try this site

http://latex.codecogs.com/editor.php

perfect math every time!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Thank you! Very useful!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi;

You are welcome and happy latexing.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

Hm, then the solution for n people seems to be 66*(2^n-2)/(12^6).

And thank you both for showing your methods.

Here lies the reader who will never open this book. He is forever dead.

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