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#1
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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. |
#2
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2^10 (2^no_of_games)
Equal to the number of different binary numbers up to 10 digits. Regards, Bernd |
#3
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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 |
#4
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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 |
#5
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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 |
#6
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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 |
#7
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David, Thanks for your help. Regards, Bernd
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#9
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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 |
#10
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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 |
#11
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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 |
#12
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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 |
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