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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: 90,654

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,654

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,654

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

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: 90,654

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

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: 90,654

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.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

Would you mind sharing?

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: 90,654

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.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

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: 90,654

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

Surely you recognize that?!

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

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: 90,654

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

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

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: 90,654

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,654

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?

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,654

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!

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,654

Hi;

You are welcome and happy latexing.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

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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