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#1 Re: Help Me ! » Combinatorics and unique pairs with a twist » 2010-10-26 11:19:28

Sure,
My data set is really (1,1,1,1) (2,2,2,2) (3,3,3,3) (4,4,4,4) (5,5,5,5) (6,6,6,6)

So I must pair each number with another unique number...for example 12, 13, 14, 15, 16...however, I can use only 4 1s so I have to drop one number, so I just chose 16 but it doesn matter which one I drop so long as  use up to 4 numbers without replicating a pair.


 


for the purposes of this excerise  my numbers are
12,13,14,15
23,24,25
34,35
46
56

I used all the 1s in available 12,13,14,15
I used all the 2s available 12,23,24,25
I used all the 3s in 13,23,34,35
I used all the 4s, 14,24,34,46
I used 3 5s 15,25,56
I used 1 6 56

This gives me 11 unique numbers without using a number more than 4 times. but as you can see I there is no way to use the last 5 or the other 3 6s...

As I was saying I am understant that I how to get the 15 total unique combinations, but how can I calculate the 4 (5,6,6,6) not being used.
This would also happen on larger sets at well depending on teh number of datasets and how they were populated. There's go to be a formula.

#2 Help Me ! » Combinatorics and unique pairs with a twist » 2010-10-26 08:28:48

jngrim
Replies: 3

Here is the problem  I have a dataset =1,2,3,4,5,6
If each number can only be used a maximum of 4 times, what is the maximum number of 2 digit and unique pairs can I create.
?
I know that if c(6,2)=6!/(2!(6-2)!)=15 but that would mean I used each number 5x's. after writing out the answer manually  I know that the the answer is 11.
I know I cant get the following numbers 16,26,36 45 which is 4, so 15-4=11 but I don't know how to maximize and exclude pairs with digits over 5. Can someone help me understand how to create this formula and create it?


Thanks alot

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