Help with combinatorics

Haidi · 2017-02-01 09:52:57

Hi there!

I need some help to solve the following:

- First off, I have to find out how many ways I can select four items from the set of 6 (repetition not allowed, order doesn't matter), so this is easy, that's 15 combinations
- Then, I need to see how many ways I can choose 6 sets out of 15, but so that, in the end, these 6 chosen sets together contain exactly 6 times item 1, 6 times item 2, 6 times item 3,... 6 times item 6.
Hope it's clear...

I assume many will find this pretty easy. I would much appreciate a (detailed) explanation rather than just a final numerical answer though.

Many thanks for whatever assistance you're able to provide.

bobbym · 2017-02-01 11:40:22

Your multi - set of choices is {1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6}? Is that right?

haidi · 2017-02-01 22:44:30

Hi bobbym,

Thanks for your prompt reply!

There was a typo in my previous post, I need to choose 9 sets out of 15 (instead of 6; obviously, it is not possible to choose 6 sets (of four items) which would contain each item six times in the end).

So, the sets (of 4) are fixed after the first step.
These are ("1" representing the item 1, "2" representing the item 2, etc.):

1234
1235
1236
1245
1246
1256
1345
1346
1356
1456
2345
2346
2356
2456
3456

Now, I need to see how many ways I can choose 9 sets to have each item six times in the end.
For instance, this is one solution:

1234
1235
1246
1256
1345
1456
2346
2356
3456

Hope it's a bit clearer.

Thank you very much for your help!

bobbym · 2017-02-01 23:24:00

I notice that there are no choices like 1135? That is not allowed?

haidi · 2017-02-02 01:56:45

No, repetition is not allowed.

haidi · 2017-02-02 05:27:59

Here's what I've come up with:

-Firstly, based on these 15 sets of four, there is a distribution pattern:
item 1 - position 1 (10)
item 2 - position 1 or 2 (4+6)
item 3 - position 1 or 2 or 3 (1+6+3)
item 4 - position 2 or 3 or 4 (3+6+1)
item 5 - position 3 or 4 (6+4)
item 6 - position 4 (10)

So, items 1 and 6 never change their position in any set of four.

- Secondly, possible positions for items 2-5 can be:
24 / 42
33
123 / 132 / 231 / 213 / 312 / 321

Thirdly, taking into account the fixed positions in the first step, all possible positions are:
item 1 - 1
item 2 - 24 / 42 / 33
item 3 - 123 / 132 / -42 / -33
item 4 - 231 / 321 / 24- / 33-
item 5 - 24 / 42 / 33
item 6 - 4

* e.g. distribution of items 4 (6 in total, in 9 sets of four), based on its position in each set (a set has four possible places), can be, say, "24-" meaning 2 items in the second place, 4 items in the third place, none in the fourth place (and none in the first place)

Now, please tell me if this makes sense...
I forgot how to use formulas, high school was a long time ago...

bobbym · 2017-02-02 05:48:01

It would only make sense to me if you could put it all together and use it to compute the number.

So far, I am unable to do that...

haidi · 2017-02-02 11:06:38

Fair enough
Though, I wouldn't have asked for help if I were able to figure it out on my own.

So, putting it all together, there are two patterns:

6---
24--
123-
-321
--42
---6

and

6---
33--
-42-
-24-
--33
---6

Additionally, the first pattern can be changed into:

6---
24--
132-
-231
--42
---6

Other combinations are not possible, if we take into account the distribution from my previous post, and the fact that the sum of each column in a pattern has to be 9.

In the end, I simply look for 9 sets out of 15 that actually fit the pattern (at least this was easy, I'm good at Sudoku ).
So, the "computed" number is 3. Maybe there are many more, maybe this has nothing to do with the correct answer though...

And these combinations are:

1234
1235
1246
1256
1345
1456
2346
2356
3456

1235
1236
1245
1346
1356
1456
2345
2346
2456

1234
1235
1236
1256
1346
1456
2345
2456
3456

Does it make sense now?
Any other suggestions?

By the way, thank you again for your interest and prompt replies.

bobbym · 2017-02-02 11:54:46

Sorry, I did not see any of that. To churn through all the possibilities will require a computer but I am not getting any ideas on how to write the program.

haidi · 2017-02-02 21:02:48

OK, I'm pretty sure this can work like this, using my patterns, just need to check once again I've got all the patterns, then it's relatively easy to find the combinations.

Thanks again for your help.

If someone else finds a neat way to solve this in the meantime, please do let me know.

bobbym · 2017-02-03 04:52:45

Hi;

Yes, I will post it right here. If you find something else please post it.

haidi · 2017-02-03 05:47:12

All right, thanks!

bobbym · 2017-02-03 12:58:47

Hi;

Welcome to the forum.

Math Is Fun Forum

#1 2017-02-01 09:52:57

Help with combinatorics

#2 2017-02-01 11:40:22

Re: Help with combinatorics

#3 2017-02-01 22:44:30

Re: Help with combinatorics

#4 2017-02-01 23:24:00

Re: Help with combinatorics

#5 2017-02-02 01:56:45

Re: Help with combinatorics

#6 2017-02-02 05:27:59

Re: Help with combinatorics

#7 2017-02-02 05:48:01

Re: Help with combinatorics

#8 2017-02-02 11:06:38

Re: Help with combinatorics

#9 2017-02-02 11:54:46

Re: Help with combinatorics

#10 2017-02-02 21:02:48

Re: Help with combinatorics

#11 2017-02-03 04:52:45

Re: Help with combinatorics

#12 2017-02-03 05:47:12

Re: Help with combinatorics

#13 2017-02-03 12:58:47

Re: Help with combinatorics

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