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September 20th, 2015, 07:14 PM   #1
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Ok so say I have 40 numbers set as 4 lines of 10...

(1-10,11-20,21-30 and 31-40). How do we work out every combination possible to get 8 numbers while using a maximum of 5 numbers from each of the 4 lines. As 8 numbers will be chosen this means that on average 2 numbers from each set of 10 will be used to make up the 8 so if we set a maximum of 5 from each set of 10 then we should have it covered. I need to know every possible combination so what I mean is usually if I were to do it manually with all 40 numbers it will be 1,2,3,4,5,6,7,8 then 1,2,3,4,5,6,7,9 then 1,2,3,4,5,6,7,10... all the way to 1,2,3,4,5,6,7,40 then start again with 1,2,3,4,5,6,8,9 all the way to 1,2,3,4,5,6,8,40 and so on but what I want to know is how can we do it and how many possiblitlies will it take if we had a maximum of 5 to choose from in each line such as 1,2,3,4,5,11,12,13.... thank you in advance for your answers
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September 21st, 2015, 01:53 AM   #2
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One way is as follows:
Get the partitions into four terms of 8, each term at most 5 (0 included) (in descending order), so
5 + 3 + 0 + 0 = 4 + 3 + 1 + 0 = ... = 4 + 2 + 1 + 1 = ... = 2 + 2 + 2 + 2.

Then for every such partition, compute how many sets you can find. Then times that by the amount of permutations of a partition.

So you start for 5 + 3 + 0 + 0 with binomial(10, 5) * binomial(10, 3) * binomial(10, 0)^2.
5 + 3 + 0 + 0 can be rearranged in (4! / (1! * 1! * 2!) = 12) ways, so times that by 12.
Okay?
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September 21st, 2015, 03:37 PM   #3
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Sorry can you please simplify this for me and tell me how many options and how to do it. Thanks.
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September 23rd, 2015, 02:24 AM   #4
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How could I simplify this for you? Could you tell me what you don't understand?
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