wu :: forums
« wu :: forums - probability and odds »
Welcome, Guest. Please Login or Register.
May 4th, 2024, 4:25pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   easy (Moderators: Icarus, william wu, SMQ, ThudnBlunder, towr, Eigenray, Grimbal)
   probability and odds		 « Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: probability and odds  (Read 2900 times)
x2862
Newbie
*





   


Posts: 10
probability and odds  
« on: Feb 6th, 2011, 8:08pm »		 Quote Quote Modify Modify
Suppose you choose n numbers from 1 to x. If k numbers are drawn from 1 to x (k > n),  what is the probability that those k numbers contain your chosen set of n numbers?
 
Now suppose you choose p sets of n numbers from 1 to k. Again if  k numbers are drawn from 1 to x (k > n), what is minimum value of p so that the probability that those k numbers contain at least 1 of your chosen sets of n numbers is 100%?
 
I figure the odds to be C(x - n, k - n) to C(x, k) - C(x - n, k - n), where C(x, n) = x! / (n! (x - n)!).
I also figure the minimum value of p to be C(x,n)/C(k,n) rounded uo to the nearest whole number.
Is this correct?
Is there an easy way to calculate the sets of p numbers for any values of x, k and, n without the use of a computer?
IP Logged
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
 Posts: 13730
Re: probability and odds  
« Reply #1 on: Feb 6th, 2011, 10:08pm »		 Quote Quote Modify Modify
on Feb 6th, 2011, 8:08pm, x2862 wrote:
Suppose you choose n numbers from 1 to x. If k numbers are drawn from 1 to x (k > n),  what is the probability that those k numbers contain your chosen set of n numbers?
C(k,n) / C(x,n). The number of ways to pick n numbers from the superset of k numbers, divided by the number of ways to pick n numbers from the whole range of x.
IP Logged
Wikipedia, Google, Mathworld, Integer sequence DB
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
 Posts: 13730
Re: probability and odds  
« Reply #2 on: Feb 7th, 2011, 10:52am »		 Quote Quote Modify Modify
Heh, the second part is easier than I though.  
p = C(x,n) - C(k,n) + 1; i.e. one more than the number possible sets of n numbers which are not subsets of the k numbers.
IP Logged
Wikipedia, Google, Mathworld, Integer sequence DB
SMQ
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
 Posts: 2084
Re: probability and odds  
« Reply #3 on: Feb 7th, 2011, 11:31am »		 Quote Quote Modify Modify
on Feb 7th, 2011, 10:52am, towr wrote:
Heh, the second part is easier than I thought.

No, I don't think it is; the formula you give is an upper bound, but p can often be substantially lower.  For instance, with x = 6, k = 3, n = 2, the minimum p is easily determined by hand to be 7, where your formula gives 12.
 
--SMQ
IP Logged
--SMQ
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
 Posts: 13730
Re: probability and odds  
« Reply #4 on: Feb 7th, 2011, 11:34am »		 Quote Quote Modify Modify
Oh wait, I didn't notice that you get to choose the sets with n numbers.
My number is for when you only get to pick p.
« Last Edit: Feb 7th, 2011, 11:39am by towr » IP Logged
Wikipedia, Google, Mathworld, Integer sequence DB
x2862
Newbie
*





   


Posts: 10
Re: probability and odds  
« Reply #5 on: Feb 7th, 2011, 4:09pm »		 Quote Quote Modify Modify
Suppose x = 9 and k = 8
Combinations without repetition (x=9, k=8)
C(9,8)=9
{1,2,3,4,5,6,7,8} {1,2,3,4,5,6,7,9} {1,2,3,4,5,6,8,9}  
{1,2,3,4,5,7,8,9} {1,2,3,4,6,7,8,9} {1,2,3,5,6,7,8,9}
{1,2,4,5,6,7,8,9} {1,3,4,5,6,7,8,9} {2,3,4,5,6,7,8,9}  
 
that's 9 combinations
 
Now lets say I picked 3 numbers.
 
Combinations without repetition (x=9, n=3)
C(9,3)=84
{1,2,3} {1,2,4} {1,2,5} {1,2,6} {1,2,7} {1,2,8} {1,2,9} {1,3,4} {1,3,5} {1,3,6} {1,3,7} {1,3,8} {1,3,9} {1,4,5} {1,4,6} {1,4,7} {1,4,8} {1,4,9} {1,5,6} {1,5,7} {1,5,8} {1,5,9} {1,6,7} {1,6,8} {1,6,9} {1,7,8} {1,7,9} {1,8,9} {2,3,4} {2,3,5} {2,3,6} {2,3,7} {2,3,8} {2,3,9} {2,4,5} {2,4,6} {2,4,7} {2,4,8} {2,4,9} {2,5,6} {2,5,7} {2,5,8} {2,5,9} {2,6,7} {2,6,8} {2,6,9} {2,7,8} {2,7,9} {2,8,9} {3,4,5} {3,4,6} {3,4,7} {3,4,8} {3,4,9} {3,5,6} {3,5,7} {3,5,8} {3,5,9} {3,6,7} {3,6,8} {3,6,9} {3,7,8} {3,7,9} {3,8,9} {4,5,6} {4,5,7} {4,5,8} {4,5,9} {4,6,7} {4,6,8} {4,6,9} {4,7,8} {4,7,9} {4,8,9} {5,6,7} {5,6,8} {5,6,9} {5,7,8} {5,7,9} {5,8,9} {6,7,8} {6,7,9} {6,8,9} {7,8,9}  
 
As we can see at least one of our 84 number sets match every set of 8 numbers. So we can say p = 84 but this is not the lowest value. I can see that we only need at most 2 sets {1,2,3) and {7,8,9}. That tells me that 2 is the lowest value for p.
Another way to get 2 is to take C(k=8,n=3) which equals 56. There are 56 sets of three numbers in a set of 8 numbers. So 84/56=1.5 rounded up that’s 2.
I still arrive at  C(x,n)/C(k,n)
 
Going further suppose n=5 instead then C{x=9,n=5)would give 126 combinations
{1,2,3,4,5} {1,2,3,4,6} {1,2,3,4,7} {1,2,3,4,8} {1,2,3,4,9} {1,2,3,5,6} {1,2,3,5,7} {1,2,3,5,8} {1,2,3,5,9} {1,2,3,6,7} {1,2,3,6,8} {1,2,3,6,9} {1,2,3,7,8} {1,2,3,7,9} {1,2,3,8,9} {1,2,4,5,6} {1,2,4,5,7} {1,2,4,5,8} {1,2,4,5,9} {1,2,4,6,7} {1,2,4,6,8} {1,2,4,6,9} {1,2,4,7,8} {1,2,4,7,9} {1,2,4,8,9} {1,2,5,6,7} {1,2,5,6,8} {1,2,5,6,9} {1,2,5,7,8} {1,2,5,7,9} {1,2,5,8,9} {1,2,6,7,8} {1,2,6,7,9} {1,2,6,8,9} {1,2,7,8,9} {1,3,4,5,6} {1,3,4,5,7} {1,3,4,5,8} {1,3,4,5,9} {1,3,4,6,7} {1,3,4,6,8} {1,3,4,6,9} {1,3,4,7,8} {1,3,4,7,9} {1,3,4,8,9} {1,3,5,6,7} {1,3,5,6,8} {1,3,5,6,9} {1,3,5,7,8} {1,3,5,7,9} {1,3,5,8,9} {1,3,6,7,8} {1,3,6,7,9} {1,3,6,8,9} {1,3,7,8,9} {1,4,5,6,7} {1,4,5,6,8} {1,4,5,6,9} {1,4,5,7,8} {1,4,5,7,9} {1,4,5,8,9} {1,4,6,7,8} {1,4,6,7,9} {1,4,6,8,9} {1,4,7,8,9} {1,5,6,7,8} {1,5,6,7,9} {1,5,6,8,9} {1,5,7,8,9} {1,6,7,8,9} {2,3,4,5,6} {2,3,4,5,7} {2,3,4,5,8} {2,3,4,5,9} {2,3,4,6,7} {2,3,4,6,8} {2,3,4,6,9} {2,3,4,7,8} {2,3,4,7,9} {2,3,4,8,9} {2,3,5,6,7} {2,3,5,6,8} {2,3,5,6,9} {2,3,5,7,8} {2,3,5,7,9} {2,3,5,8,9} {2,3,6,7,8} {2,3,6,7,9} {2,3,6,8,9} {2,3,7,8,9} {2,4,5,6,7} {2,4,5,6,8} {2,4,5,6,9} {2,4,5,7,8} {2,4,5,7,9} {2,4,5,8,9} {2,4,6,7,8} {2,4,6,7,9} {2,4,6,8,9} {2,4,7,8,9} {2,5,6,7,8} {2,5,6,7,9} {2,5,6,8,9} {2,5,7,8,9} {2,6,7,8,9} {3,4,5,6,7} {3,4,5,6,8} {3,4,5,6,9} {3,4,5,7,8} {3,4,5,7,9} {3,4,5,8,9} {3,4,6,7,8} {3,4,6,7,9} {3,4,6,8,9} {3,4,7,8,9} {3,5,6,7,8} {3,5,6,7,9} {3,5,6,8,9} {3,5,7,8,9} {3,6,7,8,9} {4,5,6,7,8} {4,5,6,7,9} {4,5,6,8,9} {4,5,7,8,9} {4,6,7,8,9} {5,6,7,8,9}  
 
 
C(k=8,n=5) gives 56 number sets 126/56=2.25 for a minimum p value of  3
 
The long way we can see we only need 3 of the sets of 5 numbers to cover all sets of 8 numbers {1,2,3,4,5}, {1,2,3,4,6} and {5,6,7,8,9}.
 
Again correct me if I am wrong as this is all new math for me.
« Last Edit: Feb 7th, 2011, 4:10pm by x2862 » IP Logged
x2862
Newbie
*





   


Posts: 10
Re: probability and odds  
« Reply #6 on: Feb 7th, 2011, 5:44pm »		 Quote Quote Modify Modify
on Feb 7th, 2011, 11:31am, SMQ wrote:

For instance, with x = 6, k = 3, n = 2, the minimum p is easily determined by hand to be 7, where your formula gives 12.
 
--SMQ

with x = 6, k = 3, n = 2,
C(x,n)=15
{1,2} {1,3} {1,4} {1,5} {1,6} {2,3} {2,4} {2,5} {2,6} {3,4} {3,5} {3,6} {4,5} {4,6} {5,6}  
 
C(x,k)=20
{1,2,3} {1,2,4} {1,2,5} {1,2,6} {1,3,4} {1,3,5} {1,3,6} {1,4,5} {1,4,6} {1,5,6} {2,3,4} {2,3,5} {2,3,6} {2,4,5} {2,4,6} {2,5,6} {3,4,5} {3,4,6} {3,5,6} {4,5,6}  
 
C(k,n)=3
{1,2} {1,3} {2,3}  
 
15/3=5 so the minimum value of p should be 5 if my formula was correct.
The actual value of p is 6 so my formula does not work after all.
 
{1,2} {1,3} {2,3} {4,5} {4,6} {5,6}
 
what i did notice is that there are exactly 4 duplicates for any givin set of 2 numbers with C(x,k).
and 20/4 is also 5. however the minimum p I get by hand is still 7.
« Last Edit: Feb 7th, 2011, 5:58pm by x2862 » IP Logged
SMQ
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
 Posts: 2084
Re: probability and odds  
« Reply #7 on: Feb 9th, 2011, 11:01am »		 Quote Quote Modify Modify
By exhaustive computer search:
 
x = 0
k = 0 1

x = 1
k = 0 1
k = 1 1 1

x = 2
k = 0 1
k = 1 1 2
k = 2 1 1 1

x = 3
k = 0 1
k = 1 1 3
k = 2 1 2 3
k = 3 1 1 1 1

x = 4
k = 0 1
k = 1 1 4
k = 2 1 3 6
k = 3 1 2 2 4
k = 4 1 1 1 1 1

x = 5
k = 0 1
k = 1 1 5
k = 2 1 4 10
k = 3 1 3 4 10
k = 4 1 2 2 3 5
k = 5 1 1 1 1 1 1

x = 6
k = 0 1
k = 1 1 6
k = 2 1 5 15
k = 3 1 4 6 20
k = 4 1 3 3 6 15
k = 5 1 2 2 2 3 6
k = 6 1 1 1 1 1 1 1

x = 7
k = 0 1
k = 1 1 7
k = 2 1 6 21
k = 3 1 5 9 35
k = 4 1 4 5 12 35
k = 5 1 3 3 5 7 21
k = 6 1 2 2 2 3 4 7
k = 7 1 1 1 1 1 1 1 1

x = 8
k = 0 1
k = 1 1 8
k = 2 1 7 28
k = 3 1 6 12 56
k = 4 1 5 7 ? 70
k = 5 1 4 4 8 ? 56
k = 6 1 3 3 4 6 ? 28
k = 7 1 2 2 2 2 3 4 8
k = 8 1 1 1 1 1 1 1 1 1
IP Logged
--SMQ
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
 Posts: 13730
Re: probability and odds  
« Reply #8 on: Feb 10th, 2011, 1:53pm »		 Quote Quote Modify Modify
I predict those three question marks (in bold) are 4, 11, 14, 20, 12, 7
For the 'row' underneath there's the sequence http://oeis.org/A066019
But it doesn't gives a new perspective for calculating those 'rows'.
« Last Edit: Feb 10th, 2011, 1:53pm by towr » IP Logged
Wikipedia, Google, Mathworld, Integer sequence DB
x2862
Newbie
*





   


Posts: 10
Re: probability and odds  
« Reply #9 on: Feb 10th, 2011, 1:57pm »		 Quote Quote Modify Modify
@SMQ, Thanks for the triangles. I found the useful even though I do not fully understand them. I notice that each triangle has all the minimum values for p given any x,k, and n.(correct me if I am wrong) Also if you stack the triangles into a pyramid one face will be Pascal’s triangle. As to how to compute the minimum value of p I still don’t see it. I f you see something I don’t I would like to know as I an very curious.(that, and the fact I like learning new things).
 
@everyone
What I have started to do in the mean time is calculate all the possibilities for small sets up to x=10.  In doing so I have made some interesting observations.
Given x=6 k=3 n=2 the minimum value of p =6
C(x,n) = 15 i.e. there are 15 ways to pull 2 numbers at random from a set of 6.
First I tried all 3,003 possible groupings of 5 and found no match.
Then I took C(15,p) = 5,005 i.e. there is 5,005 ways to group the 15 pairs into sub groups of 6.
Of those 5,005 sub groups there is only 10 sub groups that will guarantee a match within 3 numbers selected at random.
Those sets are as follows:  
{{1,2} {1,3} {2,3} {4,5} {4,6} {5,6}}
{{1,2} {1,4} {2,4} {3,5} {3,6} {5,6}}
{{1,2} {1,5} {2,5} {3,4} {3,6} {4,6}}
{{1,2} {1,6} {2,6} {3,4} {3,5} {4,5}}
{{1,3} {1,4} {2,5} {2,6} {3,4} {5,6}}
{{1,3} {1,5} {2,4} {2,6} {3,5} {4,6}}
{{1,3} {1,6} {2,4} {2,5} {3,6} {4,5}}
{{1,4} {1,5} {2,3} {2,5} {3,5} {4,6}}
{{1,5} {1,6} {2,3} {2,4} {3,4} {5,6}}
 
What I notice is there is exactly 2 of every integer 1 thru 6 in each set.
I believe I have found a way to calculate the number of times each integer should  be replicated and where. I not sure though.
 
So to test this I looked up the minimum p from the triangles you provided for x=7 k=3 and n=2. It should be 9. I then applied my algorithm and came up with the following set:
{1,2} {1,3} {1,4} {2,3} {2,4} {3,4} {5,6} {5,7} {6,7}
 
This set works for any given 3 numbers out of 7. I think I might have gotten lucky though. As my algorithm only seems to work if x=2. I will look into it further. Possibly I have to exponentially increase my algorithm. But I do not think much is possible beyond calculating the odds, combinations, and minimum p. Finding an algorithm  to calculate the actual sets without using a brute force method on large numbers would imply that P = NP.
IP Logged
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
 Posts: 4489
Re: probability and odds  
« Reply #10 on: Feb 10th, 2011, 2:24pm »		 Quote Quote Modify Modify
on Feb 10th, 2011, 1:57pm, x2862 wrote:
Finding an algorithm  to calculate the actual sets without using a brute force method on large numbers would imply that P = NP.

Per Aspera Ad Astra  Wink
« Last Edit: Feb 12th, 2011, 3:32pm by ThudnBlunder » IP Logged
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
x2862
Newbie
*





   


Posts: 10
Re: probability and odds  
« Reply #11 on: Feb 10th, 2011, 3:25pm »		 Quote Quote Modify Modify
on Feb 10th, 2011, 1:53pm, towr wrote:
I predict those three question marks (in bold) are 4, 11, 14, 20, 12, 7
For the 'row' underneath there's the sequence http://oeis.org/A066019
But it doesn't gives a new perspective for calculating those 'rows'.

 
 
Thanks for that interesting find. I am going tomorrow to the library over at cal state San Bernardino. they have the book CRC Handbook of Combinatorial Designs, which was mentioned in your link to the Encyclopedia of Integer Sequences.
 
Feel free to suggest any other books I might find helpful on the subject, as I will want to read those too.
IP Logged
SMQ
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
 Posts: 2084
Re: probability and odds  
« Reply #12 on: Feb 10th, 2011, 4:16pm »		 Quote Quote Modify Modify
on Feb 10th, 2011, 1:57pm, x2862 wrote:
I notice that each triangle has all the minimum values for p given any x,k, and n.(correct me if I am wrong)

Correct.
Quote:
Also if you stack the triangles into a pyramid one face will be Pascal’s triangle.

Also correct; that would be the n = k face.  For that case p is simply C(k,x) since you need to pick all k-sets for full cover.
Quote:
As to how to compute the minimum value of p I still don’t see it.

Nor do I, and towr's links would seem to indicate that it's an open problem since the sequences in OEIS contain unknown values.
Quote:
I f you see something I don’t I would like to know as I an very curious.

From studying the results, I believe can calculate p in the following cases:
n = 0 --> p = 1
n = 1 --> p = x - k + 1
n = 2 --> p = (x mod (k - 1))C(2,x/(k-1)) + (k - 1 - (x mod (k - 1)))C(2,x/(k-1))
n = k --> p = C(k,x)
k = x - 1 --> p = x/(x-n)
k = x --> p = 1
 
The only particularly interesting one of those is the n = 2 case (for which there may be a better closed-form expression).  My insight being that n = 2 can be viewed as an undirected graph where the elemnts of X are the vertexes and the p sets of two elements are the edges.  In this case the most efficient solution is to divide the graph into k-1 fully-connected independent subgraphs.  That way at most k-1 elements can be chosen such that no two are connected by an edge of the graph; the k-th vertex must be connected to some other already-chosen vertex by an edge, and so any set of k vertexes will contain at least one edge.
 
This idea can be extended to the cases where n = a, k = (a-1)b + 1 for any integers a and b >= 1, but I don't see a simple extension to any other cases.
 
--SMQ
IP Logged
--SMQ
SMQ
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
 Posts: 2084
Re: probability and odds  
« Reply #13 on: Feb 11th, 2011, 8:54am »		 Quote Quote Modify Modify
on Feb 10th, 2011, 1:53pm, towr wrote:
I predict those three question marks (in bold) are 4, 11, 14, 20, 12, 7
For the 'row' underneath there's the sequence http://oeis.org/A066019
But it doesn't gives a new perspective for calculating those 'rows'.

Actually, that might be an interesting result since the sequences you link to are answering a different--though related--question!
 
The problem here asks: what is the minimum number of n-sets you can pick such that every k-set contains at least one of the chosen n-sets?
 
The covering problem asks: what is the minimum number of k-sets you can pick such that every n-set is contained in at least one of the chosen k-sets?
 
It would appear, then, that p(x, k, n) = C(x,  x - n, x - k).  I guess that's not a deep observation, since if A, B X then A B BC AC.  Ah well.
 
--SMQ
« Last Edit: Feb 11th, 2011, 8:55am by SMQ » IP Logged
--SMQ
x2862
Newbie
*





   


Posts: 10
Re: probability and odds  
« Reply #14 on: Feb 11th, 2011, 11:10pm »		 Quote Quote Modify Modify
I have found the following which is interesting
 
http://oeis.org/A084546
 
The numbers match C(C(x,n),p) so if I could produce those same results without p then I could easily calculate p.
 
I also notice that that the lowest value p could be in this case would be the value just before the median value. I still am looking into wither this can be of any use to finding p.
IP Logged
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
 Posts: 13730
Re: probability and odds  
« Reply #15 on: Feb 12th, 2011, 6:31am »		 Quote Quote Modify Modify
In what way does that sequence has anything to do with this problem? I don't see our numbers match up in any way with that sequence. ?!
IP Logged
Wikipedia, Google, Mathworld, Integer sequence DB
x2862
Newbie
*





   


Posts: 10
Re: probability and odds  
« Reply #16 on: Feb 12th, 2011, 2:46pm »		 Quote Quote Modify Modify
Perhaps that sequence dose not relate to the problem at hand. perhaps it does. I have no idea. But I noticed some interesting things.
Here is what I noticed, given x=6, k=3, and n = 2 what are the possible ways to arrange the 15 different sets of 2 into sets of p.  
If p = 1 then C(15,1) =    15.
If p = 2 then C(15,2) =    105.  
If p = 3 then C(15,3) =    455.
If p = 4 then C(15,4) =    1365.
If p = 5 then C(15,5) =    3003.
If p = 6 Then C(15,6) =   5005. //here is our minimum p (6). It is also the sixth value.
If p = 7 Then C(15,7) =   6435. //here is the apex of the curve.
If p = 8 Then C(15,8) =   6435.
If p = 9 Then C(15,9) =   5005.
If p = 10 Then C(15,10) = 3003.
If p = 11 Then C(15,11) = 1365.
If p = 12 Then C(15,12) = 455.
If p = 13 Then C(15,12) = 105.
If p = 14 Then C(15,14) = 15.
 
Notice the matching numbers?  
15, 105, 455, 1365, 3003, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15
 
Now lets change the problem to x=7, k=3, n=2.
Now there are 21 different ways to arrange 7 numbers into groups of two
 
Starting with 21 the sequence gives  
21, 210, 1330, 5985, 20349, 54264, 116280, 203490, 293930, 352716, 352716, 293930, 203490, 116280, 54264, 20349, 5985, 1330, 210, 21
 
Doing the above for p I find the same thing.
 
If p = 1 Then C(21,1) =   21
If p =  2 Then C(21,2) =   210
If p =  3 Then C(21,3) =   1330  
If p =  4 Then C(21,4) =   5985  
If p =  5 Then C(21,5) =   20349  
If p =  6 Then C(21,6) =   54264  
If p =  7 Then C(21,7) =   116280  
If p =  8 Then C(21,8) =   203490  
If p =  9 Then C(21,9) =   293930 //here is our minimum p (9).  
If p =  10 Then C(21,10) =   352716 //here is the apex of the curve
If p =  11 Then C(21,11) =   352716  
If p =  12 Then C(21,12) =   293930  
If p =  12 Then C(21,13) =   203490  
If p =  14 Then C(21,14) =   116280  
If p =  15 Then C(21,15) =   54264  
If p =  16 Then C(21,16) =   20349  
If p =  17 Then C(21,17) =   5985  
If p =  18 Then C(21,18) =   1330  
If p =  19 Then C(21,19) =   210  
If p =  20 Then C(21,20) =   21
 
Hope this clarifies  what I noticed.
« Last Edit: Feb 12th, 2011, 2:47pm by x2862 » IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board