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Hi;
Yes it would...
(03 03 13) select 1 =What is the formula for calculating the number of possible combinations
I thought you might need the answer to it from right there.
well thanks but I still think you guys do not understand the problem enough to evaluate it for a solution.
The number of combinations of 3,3,13 are only one because in combinations order does not count.
There are 3 permutations of 3,3,13
Bobby,
Can you explain further as I am not following.... I understand the formula fine but please explain where the 3,3,13 comes from...
II
So you want 2 quick picks, and then you want 8 more picks that are made out of the first 2 picks somehow?
Are the other 8 picks similar to the first two in that they have 1 of the 5 the same and one of the mega ball the same?
Are we looking for the number of possible combinations using these rules?
Here are two answers.
If I want to generate the possible combination that can be made if I am using two rows to pick from as mentioned above, I already know the possible combinations will be 64. This is only possible if there are no matching number is the two rows that i am picking from...
If I want to generate the possible combination that can be made if I am using THREE rows to pick from as mentioned above, I already know the possible combinations will be 729. This is only possible if there are no matching number is the two rows that i am picking from...
This is easily achieved like this: for 2 rows the this simple formula is (2^6)=64. for 3 rows the this simple formula is (3^6)=729. For any number of rows its the same like say 12 rows to choose form would be (12^6)=2,985,984
There's only one wrong with this formula... If there is a matching set of balls in any of the columns it changes the outcome and thus I DO NOT know the formula to calculate that.
Take the number TWO (02) in the below example. If looping thru this using the above formula I could end up with 02 02 08 13 05 03. This is an invalid combination because two of the first 5 numbers match. I have to be able to eliminate those possibilities. The formula to do this would probably need to know the number of balls we are selecting from. In the case below it would probably be 14 because one of the 02's would have to be kicked out. (remembering that the sixth position is independent of the first 5 positions).
01 02 03 04 05 03
06 07 08 09 10 07
02 11 12 13 14 13
Lets say that each of the columns above is a bucket. The first five buckets get one number selected from them on each possible combination but none of them can match.
(01 06 02) select 1 =
(02 07 11) select 1 =
(03 08 12) select 1 =
(04 09 13) select 1 =
(05 10 14) select 1 =
But wait, if there are any matching numbers in any of the 5 buckets, we have to kick it out so there are no matching numbers then do the formula maybe like below in which the 02 was dropped from the first bucket (or column). Now we can still pull random numbers from each bucket and never have a matching set of numbers within the first 5 buckets. This narrows the possible combinations:
(01 06) select 1 =
(02 07 11) select 1 =
(03 08 12) select 1 =
(04 09 13) select 1 =
(05 10 14) select 1 =
Mega ball (sixth digit). Any number can be selected from this each time and can match any of the selections above to come up with six different numbers than the time before.
(03 07 13) select 1 =
What if on the Mega ball above there were a matching ball in the bucket? This would narrow the selections in this bucket to only two right? Yes it would...
(03 03 13) select 1 =
What is the formula for calculating the number of possible combinations given any number of rows to pull from and no matter how many matching numbers there are to be kicked out??
II
Hi;
So this Mega ball can be any number in that drum from 1 to 46. With replacement for the next row?
HI.
Yes the mega ball can be any number, but, it has to be taking form the rows that are given to randomize. Say I want to randomize two rows of numbers. the mega ball in each row are 12 & 06... then each combination output from the process would have to always be either the 12 or 06 and no other.... The same for the first 5 white balls. If we have two rows of numbers to work with, then all the computer numbers must be selected from those two rows AND, must remain in the same original position (or column).
II
Hi PieLover;
I am asking what the last column of numbers looks like. What the last column contains is important to solving the problem.
Once I have all the info then I can work on the problem.
Hi bobby
Its just lottery numbers where the mega ball is the last ball and is drawn out of a totally different drum. They currently range from 1-46. The first 5 are the white balls which also are in a different drum that goes from 1-56. So on the first 5 white balls, once a number is drawn that number is gone and there only remains 55 to choose from. But I want to take a few rows and randomize them but keeping them in the same column or position. I sure hope this helps.... Thanks a million.
II
I like small examples when I can't understand a big example.
So can you make this problem easier somehow, by cutting
down on the list size and the last number is still okay I guess.
But how about a list of just 2 numbers that can be from 0 to 9
and then a special final number that can also be 0 to 9, but the
first two numbers can't be the same right? And also, what about
the column issue between say 5 or 10 of these 3-lists.
Can you explain that a little better, or make it smaller and easier
for me to work on? I just think I might be able to make some
headway if we start really small, and then later work to the
actual example you are interested in. Maybe bobbym can
handle this general case, but I frankly need smaller because
it helps me to understand what is and isn't allowed and makes
explaining the rules easier, and examples are easier to
list out in full, many times. Sorry for the suggestion if you don't
like it, I am sorry I am so dumb.
Thanks.
I have explained it pretty simple I think, its really not that hard to understand. It looks as if the thread has dead-ended... Thanks anyways.
II
Why don't you just always put the numbers in numerical order (the first 5), from lowest to highest?
Wouldn't that make the number of permutations the same as the number of combinations?
John,
I can easily place all the numbers in numerical order. If fact I will have to do this anyways to compare it quickly to the numbers that have already been selected. So yes you may be correct but does knowing this make a particular formula come to mind to solve the issue?
Hi;
So you are saying the last column can be any number from 1 to 13?
bobby,
the range of the numbers here is not relevant.
Thanks,
II
Hi;
01 07 12 12 14 13
You have tow twelves in the same row, how is that possible?
Yes thank you for pointing that out. It was a total mistake and I have corrected the post. No two numbers may match in within the first 5 balls, and the sixth number may match any one ball in any given row.
Thank You
II
Thank you for your kind reply
As an example, output would be generate from these:
01 02 03 04 05 03
06 07 08 09 10 07
02 11 12 13 14 13
and some output would be:
+++++++++++++
01 07 12 04 14 13
01 11 03 04 05 13
06 11 03 09 10 13
06 07 08 09 10 07 ----- This row of output has a 07 matching in position 2 and 6. This is perfectly fine as the 6th position is independent and can match any of the first 5.
02 07 08 13 10 03 ----- Notice in this row there could have been a 02 in position 2 as well (second column) but this isn't allow.
01 02 12 09 14 03
02 11 03 04 05 07
+++++++++++++
The numbers above would be the output within the plus signs.
Notice all balls remain in their original columns (or) positions.
none of the first five positions are allow to match, however, the sixth position is completely seperate and CAN match any one number of the first five. So i'm thinking it could be worked like two different problems then placed together as a final step.
Thank you so much
II
Math Question (Combinations)
Hi I love this website. I found an example on combination with no repeats. It helped me solve a 2 year problem I had. There is one more problem i have that relates to that one also.
For multiple rows of 6 lottery numbers, stacked lets say 2 or 3 rows high to keep it simple (seen below). Each of the first five numbers must always stay their original position or column. In other words, looking at the example below; the numbers 01 06 02 would always have to be the first number (first column). The order of the first 5 numbers doesn't matter as long as the output of each combination is different by at least one number from all the others that are calculated. The first 5 numbers can NOT match. The 6th number in the rows (03 07 13 below) is separate from the first 5 and its OK if they match any of the first 5 numbers.
I can calculate the combination if there are no numbers that match in any rows or columns (first five is separate from last column) but if there is one matching number, it narrows the combination and I cant figure out the formula for that. an example below:
The formula below won't work in this case:
n!
-----
r!(n-r)!
The above formula above works great if it doesn't matter what column the first 5 numbers fall into as long as i multiple the answer by the number of rows without matching numbers in the sixth column. In this case if there are matches like the 02 which is found in row 1 and 3, i just drop one of those off and count it as 14 objects instead of 15. At least that is the only way I can get it to work.
But in this question, I am particularly talking about keeping the numbers in the same original positions... First five can't match each other, the 6th number can match any of first 5. The resulting answers must all have at least one number different. How many combination can be calculated? What if there are 2 balls that are matching in the first 5 of the three rows, then how many combination are possible?
01 02 03 04 05 03
06 07 08 09 10 07
02 11 12 13 14 13
In the above example, the number 02 is there twice. Do you know the formula to calculate this? Remember the numbers can only be mixed with the numbers in the same column. so the number 01 would always have to come out in column 1 and likewise number 04 would always have to result into column 4. The example of the 02 being in a different column from the other 02 would even be a different problem if it were in the same column but in row 3 wouldn't it? That would mean the column with the two 02's would actually have only one combination. It is confusing me quite a bit. Can you shed some insight on this please? I know I am close but I just can't quiet get it.
Best Regards,
Pie II
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