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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

In a Diagram shown in figure, Not that the Grid path from **F** to **G** is missing.

So path from **A** to **B** can not pass from **F** and **G**.

How many path are there from **A** to **B**

assume all path only have steps going up or to the right.

*Last edited by jacks (2013-09-29 05:30:44)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

Where is D and E?

**In mathematics, you don't understand things. You just get used to them.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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

Sorry Bobbym , actually it is **A** and **B**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

Hi;

I am getting 38 ways.

**In mathematics, you don't understand things. You just get used to them.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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

Would you like to explain it to me, Thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

For ones that have missing intersections and are small the vertex counting method will work fine.

The drawing shows the number of paths to each node.

Each node is the sum of the two nodes underneath and to left of it.

**In mathematics, you don't understand things. You just get used to them.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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

Thanks Bobbym Got it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

There is an analytical answer for these type but the counting method is easier to understand.

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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

Would you like to explain me analytical answer.

I also tried for that method but could not get it.

(But I like yours Counting method.)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

Hi;

By combinatorics:

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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

Thanks Bobbym would you like to explain me to it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

It is just adding up all the paths in the lattice around the missing intersection.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Hi

is also a way.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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

To anonimnystefy I have understand

(Which is Total no. of path from **A** to **B**, If There is a connectivity between F to G)

but Did not understand

Would you explain it to me.

Thanks

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

It's the number of paths from A to F times the number of paths from G to B.

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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

To Admin (Bobbym) i did not understand it

It is just adding up all the paths in the lattice around the missing intersection.

Thanks

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

I think I know which points he took, but I cannot be sure.

I think he used (2,3), (3,0) and (3,1), A being (0,0) and B being (5,4).

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: 82,779

Hi;

Look at my drawing in post #6.

I picked (2,3), (3,1) and (4,0) and calculated the paths from there.

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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

It's the number of paths from A to F times the number of paths from G to B.

To anonimnystefy I did not understand it

why we minus it.

please explain it to me

Thanks

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

anonimnystefy wrote:

It's the number of paths from A to F times the number of paths from G to B.

This is the reason. These are all the paths that become unavailable when we remove the line.

Hi bobbym

Yes, I meant (4,0).

*Last edited by anonimnystefy (2013-09-30 07:59:54)*

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: 82,779

Hi;

I did not see your post 17, I was answering his question.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Even better then.

I guess the question is fully answered now. Three ways is enough.

I just want to try to do one more thing.

*Last edited by anonimnystefy (2013-09-30 08:04:30)*

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: 82,779

There is another more general way. That maybe can answer for many missing intersections.

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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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

Thanks Admin and anonimnystefy got it

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,779

Hi;

You are wlecome.

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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