combinations
 

Hello, People!
I am looking for a formula that will help me determine how many different
outcome I will get.

I have illusrtated below a simple format the way it will be layed out.

Home Visitor
1. Tor vs Phil
2. NYI vs NYR
3. Pitts vs NJ
4. FLA vs TB
5. Cal vs EDM
6. MTL vs Ott
7. Bos vs Det
8. Buf vs LA
9. Van vs Nash
10. NC vs Min

Please note the formula that I am looking for: In game #1 there are two
possible outcomes . Either home team wins or Visitor. The two teams cannot
play the following teams below them or above them. Can any one help me or
guide me to a formula, hopefully with step by step instructions on how to
perform this. I would really appreciate this.

combinations
 

2^10 (2^no_of_games)
Equal to the number of different binary numbers up to 10 digits.

Regards,
Bernd

combinations
 

Thanks Bernd!
Could you provide a little description. I am a little confused on where and
how to input the formula.

Regards

"Bernd" wrote:

2^10 (2^no_of_games)
Equal to the number of different binary numbers up to 10 digits.

Regards,
Bernd

combinations
 

Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
....
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

Bernd,

Your link return a *Page not found* for me
--
Regards,

Sandy
In Perth, the ancient capital of Scotland
and the crowning place of kings


with @tiscali.co.uk


"Bernd" wrote in message
ups.com...
Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
...
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

I think you've only looked at half of the url, Sandy. It is:
http://www.sulprobil.com/html/longdec2bin__.html
--
David Biddulph

"Sandy Mann" wrote in message
...

Bernd,

Your link return a *Page not found* for me

"Bernd" wrote in message
ups.com...
Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
...
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

David, Thanks for your help. Regards, Bernd

combinations
 

Thank you David. The link did not seem to be far enough along for it to
have been hit by a line wrap but then there no other*break-point* Also the
full colon after the 2nd html fooled me into thinking that that line
referred to the text below it.

--
Regards,

Sandy
In Perth, the ancient capital of Scotland
and the crowning place of kings


with @tiscali.co.uk


"David Biddulph" wrote in message
...
I think you've only looked at half of the url, Sandy. It is:
http://www.sulprobil.com/html/longdec2bin__.html
--
David Biddulph

"Sandy Mann" wrote in message
...

Bernd,

Your link return a *Page not found* for me

"Bernd" wrote in message
ups.com...
Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
...
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

Hi Bernd --

Question going beyond your formula -- I am trying to find an additional
formula that looks at how many combinations of 2 digits or 3 digits or so
forth -- so if I had something where I had 5 things then I have 2^5-1 (I
don't use the 00000 in this case) 31 different combos. What I am trying to
figure out is that I know I have 31 combos, but the 31 combos come from 5
single digits (00001,00010,00100,01000,10000) and a bunch come from 2 digits
(00011,00101, etc). I am trying to figure out a formula for determining how
many 2 digits combos are possible, how many 1 digits are possible, 3 digits,
etc.

If you know off the top of your head I would appreciate the help --

Thank you --

Jim

"Bernd" wrote:

Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
....
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

Rubble - Check out Pascal's Triangle. Your answers are in the sixth row.

Alternately,
=COMBIN(5,1)
=COMBIN(5,2)
=COMBIN(5,3)
..
..
..
- David Hilberg

Rubble wrote:
Hi Bernd --

Question going beyond your formula -- I am trying to find an additional
formula that looks at how many combinations of 2 digits or 3 digits or so
forth -- so if I had something where I had 5 things then I have 2^5-1 (I
don't use the 00000 in this case) 31 different combos. What I am trying to
figure out is that I know I have 31 combos, but the 31 combos come from 5
single digits (00001,00010,00100,01000,10000) and a bunch come from 2 digits
(00011,00101, etc). I am trying to figure out a formula for determining how
many 2 digits combos are possible, how many 1 digits are possible, 3 digits,
etc.

If you know off the top of your head I would appreciate the help --

Thank you --

Jim

"Bernd" wrote:

Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
....
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

Thank you !!! That helps me a ton . . . my ability to think through
mathematical formulas has diminished greatly since college -- thank you

"David Hilberg" wrote:

Rubble - Check out Pascal's Triangle. Your answers are in the sixth row.

Alternately,
=COMBIN(5,1)
=COMBIN(5,2)
=COMBIN(5,3)
..
..
..
- David Hilberg

Rubble wrote:
Hi Bernd --

Question going beyond your formula -- I am trying to find an additional
formula that looks at how many combinations of 2 digits or 3 digits or so
forth -- so if I had something where I had 5 things then I have 2^5-1 (I
don't use the 00000 in this case) 31 different combos. What I am trying to
figure out is that I know I have 31 combos, but the 31 combos come from 5
single digits (00001,00010,00100,01000,10000) and a bunch come from 2 digits
(00011,00101, etc). I am trying to figure out a formula for determining how
many 2 digits combos are possible, how many 1 digits are possible, 3 digits,
etc.

If you know off the top of your head I would appreciate the help --

Thank you --

Jim

"Bernd" wrote:

Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
....
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd

combinations
 

You're welcome. I'm sure it all comes simply *rushing* back to you now!

- David

Rubble wrote:
Thank you !!! That helps me a ton . . . my ability to think through
mathematical formulas has diminished greatly since college -- thank you

"David Hilberg" wrote:

Rubble - Check out Pascal's Triangle. Your answers are in the sixth row.

Alternately,
=COMBIN(5,1)
=COMBIN(5,2)
=COMBIN(5,3)
..
..
..
- David Hilberg

Rubble wrote:
Hi Bernd --

Question going beyond your formula -- I am trying to find an additional
formula that looks at how many combinations of 2 digits or 3 digits or so
forth -- so if I had something where I had 5 things then I have 2^5-1 (I
don't use the 00000 in this case) 31 different combos. What I am trying to
figure out is that I know I have 31 combos, but the 31 combos come from 5
single digits (00001,00010,00100,01000,10000) and a bunch come from 2 digits
(00011,00101, etc). I am trying to figure out a formula for determining how
many 2 digits combos are possible, how many 1 digits are possible, 3 digits,
etc.

If you know off the top of your head I would appreciate the help --

Thank you --

Jim

"Bernd" wrote:

Hello,

You have 10 games with a possible result of 0 or 1,
so your result universe is
0000000000
0000000001
0000000010
0000000011
0000000100
....
1111111111
That makes 2^10. Just write into any cell
=2^10
which is 1024
Or take my UDF longbin2dec from http://www.sulprobil.com/html/
longdec2bin__.html:
=longbin2dec("10000000000")
which results in 1024 again.

Regards,
Bernd