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## #1 Re: Coder's Corner » Combinations within a Subset of Combinations » 2016-06-21 11:32:33

Probably best if I could email the Excel I created manually to show you
My brain hurts, but I "think" you may have it nailed
The trick is to have ALL the 4 Combinations contained in minimum of 6 numbers
Some 4 combinations will appear more than once but that has to be.

## #2 Re: Coder's Corner » Combinations within a Subset of Combinations » 2016-06-20 11:24:15

Can be written in anything!
I could try but I don't know what the terms are
Trust me I have looked and looked for examples
Thought I would ask here, as I always use the www link to create the "combinations"

I have an Excel that I use that displays where the 4 combinations are found in the 6 numbers
And at the end I export a CSV file of the 6 numbers

## #3 Coder's Corner » Combinations within a Subset of Combinations » 2016-06-19 21:14:24

gra0001
Replies: 7

Hi, looking for a person to program or maybe use VBA in Excel to Solve something.
Lets say I have 12 Numbers 1 to 12, so there are 495 sets of 4 combinations
Sample: Using this link (below) and Selecting 12 numbers and 4 numbers to choose
Is Order important?= No    Is Repetition allowed?=No
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

Now I want to list the minimum number of 6 number sets that contain the 4 number combinations
Manually I worked out I could do it in 42 sets of 6 numbers that contained all 495 combinations
That is we would have all the 4 combinations in a set of 6 numbers of the 12 numbers.
I don’t know what you call this Subsets?
EG The 12 Numbers are 1,2,3,4,5,6,7,8,9,10,11,12
There are 495 combinations of 4 numbers
-In say 1,2,3,4,5,6 (6 Numbers) there are 15 Combinations of 4
(1,2,3,4)(1,2,3,5)(1,2,3,6)(1,2,4,5)(1,2,4,6)(1,2,5,6)(1,3,4,5)(1,3,4,6)(1,3,5,6)(1,4,5,6)(2,3,4,5)(2,3,4,6)(2,3,5,6)(2,4,5,6)(3,4,5,6)
-In say 1,2,3,7,8,9 (6 Numbers) there are 15 combinations of 4
(1,2,3,7)(1,2,3,8)(1,2,3,9)(1,2,7,9)(1,2,8,9)(1,3,7,8)(1,3,8,9)(1,7,8,9)(2,3,7,8)(2,3,7,9)(2,7,8,9)(3,7,8,9)(1,3,7,9)(2,3,8,9)(1,2,7,8)
So in 42 Lines of sets of 6 number combinations I created a list that contain all 495 combinations of 4
42 Lines was the minimum,  and I believe the sets of 6 was balanced because I used each number in the set of 6 numbers 21 times each

The Program I need is:
Enter the Total number of Numbers EG 12 (But can be any number to 50)
Enter the Total number used in each SET EG 6 (But can be any number to 10)
Enter the Total number of Combinations to cover EG 4 (But can be any number to 10)
Then display 2 things:
-The string of the 6 Numbers (1 Row for every set of 6 numbers)
-All the Combinations of 4 and what row they appear in (for error checking)
-Note: It needs to be done in the Minimum of Rows of 6 numbers
Yes some duplication of obtain the 4 combinations may happen as it did in my 42 set of 6

## #4 Re: Help Me ! » Permutation Formula that I cant solve » 2015-08-13 09:52:52

Yes, that is what I am after.
And because the current answer all the numbers are used the same time and the spread is balanced.
That is what I am after.
I still think it can only be done manually, but then how to you check to see it is done correctly?
THanks

## #5 Re: Help Me ! » Permutation Formula that I cant solve » 2015-08-12 21:19:52

Hi,
I think it is a interesting one.
I have an Excel Spread Sheet to send if that helps
David

## #6 Re: Help Me ! » Permutation Formula that I cant solve » 2015-08-11 13:29:03

Maybe I should say combination not permutations......sorry
If they would call a combination lock a permutation lock...it would of been better for my brain

So line 2 contains a unique set of 4 combinations
Line 2 is 1,2,3,7,8,9 and there is the set of 4 combinations of 1,2,3,7 and 1,2,3,8 and 1,2,3,9 and 2,3,7,8 etc etc
These "may" not exist in any of the other lines of selected numbers of 6
The result is 6 numbers are selected from the 12
Out of those 6 numbers there are combinations of 4 that are not repeated in any other lines.
Just trying to work out a formula that is balanced so each number in the set of 6 is presented the same amount of times.

Maybe it has to be completed manually?

## #7 Re: Help Me ! » Permutation Formula that I cant solve » 2015-08-11 10:15:22

I try to explain better:
So the full 12 numbers are 1,2,3,4,5,6,7,8,9,10,11 and 12
In the reply above I gave the 42 sets of 6 numbers
In that sample each of the 12 numbers ONLY appears 21 times, so this tells me it is a balanced result.

I was to "believe" that in the example EG line 1-1,2,3,4,5,6 that this was only time 1,2,3,4 existed together.
But so does every other set of four permutations in LINE 1 (1,2,3,5 / 1,2,3,6 / 2,3,4,5 / 2,3,4,6 ...etc)
and I call this a set of 4, but because we are extracting 6 numbers and within those 6 we can cover the set of 4 permutations.

Where the normal calculation of having 12 numbers and wanting every combination of 4 is 495 Permutations.
But here we are using a string of 6 numbers but ONLY looking at sets of 4 in it.
It is hard to explain but hope that helps

1-1,2,3,4,5,6,
2-1,2,3,7,8,9,
3-1,2,3,10,11,12,
4-1,2,4,7,8,10,
5-1,2,4,9,11,12,
6-1,2,5,7,8,11,
7-1,2,5,9,10,12,
8-1,2,6,7,8,12,
9-1,2,6,9,10,11,
10-1,3,4,7,9,10,
Etc

## #8 Re: Help Me ! » Permutation Formula that I cant solve » 2015-08-10 20:35:14

Nope, not a progressive dinner.
For a project that I am working on.
A formula type game, there is a group of us that use the sample i sent you to increase our chances.
Some of the group are maths teachers...and it has their heads scratching.
But if I could understand why the person who invented it (the sample) with 12 numbers used 42 sets and each number appeared 21 times i could probably sort out how to increase the set to 7 (from 6 numbers)

## #9 Re: Help Me ! » Permutation Formula that I cant solve » 2015-08-10 18:22:37

Someone did it for me 20 Years ago.
It was 12 Numbers and there were 42 sets of 6 numbers
So every number was used 21 times
It covered the Majority of combinations and at least every set of 4 was presented
I am not sure how they got this result?
But I have a problem where I now have to have 7 numbers...it may well take more than 42 sets
Thanks David

1-1,2,3,4,5,6,
2-1,2,3,7,8,9,
3-1,2,3,10,11,12,
4-1,2,4,7,8,10,
5-1,2,4,9,11,12,
6-1,2,5,7,8,11,
7-1,2,5,9,10,12,
8-1,2,6,7,8,12,
9-1,2,6,9,10,11,
10-1,3,4,7,9,10,
11-1,3,4,8,11,12,
12-1,3,5,7,9,11,
13-1,3,5,8,10,12,
14-1,3,6,7,9,12,
15-1,3,6,8,10,11,
16-1,4,5,7,10,11,
17-1,4,5,8,9,12,
18-1,4,6,7,10,12,
19-1,4,6,8,9,11,
20-1,5,6,7,11,12,
21-1,5,6,8,9,10,
22-2,3,4,8,9,10,
23-2,3,4,7,11,12,
24-2,3,5,8,9,11,
25-2,3,5,7,10,12,
26-2,3,6,8,9,12,
27-2,3,6,7,10,11,
28-2,4,5,8,10,11,
29-2,4,5,7,9,12,
30-2,4,6,8,10,12,
31-2,4,6,7,9,11,
32-2,5,6,8,11,12,
33-2,5,6,7,9,10,
34-3,4,5,9,10,11,
35-3,4,5,7,8,12,
36-3,4,6,9,10,12,
37-3,4,6,7,8,11,
38-3,5,6,9,11,12,
39-3,5,6,7,8,10,
40-4,5,6,10,11,12,
41-4,5,6,7,8,9,
42-7,8,9,10,11,12,

## #10 Help Me ! » Permutation Formula that I cant solve » 2015-08-10 16:45:27

gra0001
Replies: 19

Can you help me with a permutations calculation
If there was a total of 12 numbers (1,2,3,4,5,6,7,8,9,10,11,12) (n)
I wanted to have a set of 7 numbers (r)
But I ONLY wanted each set to have at least the permutations of 4 numbers and the remaining 3 numbers to not be repeated in any set together but be unique.
SO if ANY of the 4 numbers selected out of the 12 was chosen it would appear in the result.