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 July 14th, 2016, 01:53 AM #1 Newbie   Joined: Jul 2016 From: Far away in a land... Posts: 2 Thanks: 0 How many possibilities do we have when... Hello, everyone! Thank you in advance for your time. I will try to describe my problem as best as I can. We have a column with 3000 rows and each row says either W or L. How many different possibilities can this create? I was thinking 2 with an exponent of 3000. Which would give pretty much infinite possibilities. (Not sure if this is even correct. It is just my assumption.) Now the hard part. What if the W makes an appearance in the range of 50-55% cases and L 45%-50%. (Of course they give only 100% together. So if W finds itself in the column 53% of the time then L has to be the other 47%.) So I was hoping there would be a mathematical way to find out how many possibilities would this give. Logic tells me there would be less possibilities because the range is now limited (eventhough its not set to just lets say 55% for W and 45% for L but anywhere from 50-55% for W and 45-50% for L. It is still only 5% of the whole spectrum of previous possibilies /or so I think/) Any thoughts on this subject are appreciated! ruth
July 14th, 2016, 06:11 AM   #2
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 Originally Posted by ruth26 Hello, everyone! Thank you in advance for your time. I will try to describe my problem as best as I can. We have a column with 3000 rows and each row says either W or L. How many different possibilities can this create? I was thinking 2 with an exponent of 3000. Which would give pretty much infinite possibilities. (Not sure if this is even correct. It is just my assumption.)
Yes, that would be $\displaystyle 2^3000$. I am not sure what you mean by "pretty much" but that is definitely not "infinite".

Quote:
 Now the hard part. What if the W makes an appearance in the range of 50-55% cases and L 45%-50%. (Of course they give only 100% together. So if W finds itself in the column 53% of the time then L has to be the other 47%.) So I was hoping there would be a mathematical way to find out how many possibilities would this give. Logic tells me there would be less possibilities because the range is now limited (eventhough its not set to just lets say 55% for W and 45% for L but anywhere from 50-55% for W and 45-50% for L. It is still only 5% of the whole spectrum of previous possibilies /or so I think/) Any thoughts on this subject are appreciated! ruth
If W appears in exactly half of the rows and L in exactly half then it becomes a matter of permuting 50 W's and 50 L's. We have 100 choices for where to put the first "W" then 49 choices for where to put the second "W", etc., down to 51 choices for where to put the last "W". Once that has been done, there are no more "choices". All of the "L"s go in the remaining rows. That is 100*99*98*...*52*51= 100!/50! ("100 factorial"- 100*99*98*...*2*1 we divide by 50!= 50*49*...*2*1 to cancel the integers we don't want).

Similarly if there are 55 Ws and 45 Ls then we have 100 choices for where to put the first "W", 99 for where to put the second "W" down to 100- 55+ 1= 46 places to put the last "W". There are 100!/45! ways to do that.

 July 14th, 2016, 06:32 AM #3 Newbie   Joined: Jul 2016 From: Far away in a land... Posts: 2 Thanks: 0 Yeah, sorry, it definitely isnt an infinite number but it being a 904 figure number is a bit too much possibilities to really go through. "If W appears in exactly half of the rows and L in exactly half then it becomes a matter of permuting 50 W's and 50 L's.We have 100 choices for where to put the first "W" then 49 choices for where to put the second" Getting a little confused here. If we have 3000 rows and its 50% for each W and L, shouldnt there be 3000 choices where to put the first W instead of the 100 mentioned? Or is it counted in percentages as in 100 represents the whole 3000 rows? Last edited by ruth26; July 14th, 2016 at 07:08 AM.
 July 14th, 2016, 01:12 PM #4 Global Moderator   Joined: May 2007 Posts: 6,540 Thanks: 591 Combination solution. n rows and k wins, no. of combinations = $\displaystyle \frac{n!}{k!(n-k)!}$. For large n, use Sterling's approximation $\displaystyle n!\approx \sqrt{2\pi n}(\frac{n}{e})^n$.

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