Sets of numbers Probability - need help

Question

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

Answer 1

We use a strategy that does not generalize nicely.

There are $\binom{40}{8}$ ways to do the job with no restrictions. Now we subtract the number of bad choices.

A bad choice has $6$, $7$, or $8$ numbers from some row. I will count the bad choices that have exactly $6$ numbers from some row, and you can deal with the other two types of bad choices.

The row from which we choose $6$ can be chosen in $\binom{4}{1}$ ways. The actual numbers can then be chosen in $\binom{10}{6}$ ways, and the other two in $\binom{30}{2}$ ways. That gives a total of $\binom{4}{1}\binom{10}{6}\binom{30}{2}$ bad choices with $6$ in some row.