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There are 4 sets of numbers

set 1: (1,2,3,4,5,6.....19) = 19 numbers
set 2: (10,11,12,13,14,15......29) = 20 numbers
set 3: (10,11,12,13,14,15......39) = 30 numbers
set 4: (20,21,22,23,24,25......39) = 20 numbers

How to calculate the total possible unique combinations from those 4 sets? No two numbers should be the same in each set. e.g. (1,10,22,35) or (15,21,25,35) etc.

Thank you.

Jack
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  • Welcome to MathSE. What have you tried? Where are you stuck? – N. F. Taussig Apr 03 '22 at 18:29
  • Are the selections sets or ordered quadruples, that is, is $(a, b, c, d)$ different from $(a, c, b, d)$ when $b \neq c$? – N. F. Taussig Apr 03 '22 at 18:37
  • @N.F.Taussig it doesn't matter. As long as they are unique. I was thinking maybe it is just simply 39 choose 4 = 82251. But not 100% sure. – Jack Apr 03 '22 at 18:45
  • If I understand your comment correctly, we are selecting subsets rather than ordered quadruples. Your guess $\binom{39}{4}$ cannot be correct since it includes sets which include two or more numbers less than $10$, which is not possible if you select only one number from each of your four sets. However, you could subtract those cases from the $\binom{39}{4}$ four-element subsets of ${1, 2, 3, \ldots, 39}$ to get your answer. – N. F. Taussig Apr 04 '22 at 00:32
  • If I understand the problem correctly, another issue is that at least one of your numbers must be at most $19$, at least two of your numbers must be at least $10$, and another number must be at least $20$. These restrictions must also be taken into account. – N. F. Taussig Apr 04 '22 at 00:42
  • @N.F.Taussig yes. each set will have a minimum and a maximum value range, as long as they are not the same when in the combination. – Jack Apr 04 '22 at 02:16
  • I do not see a nice approach. Either you need to do a lot of casework or you need to use the Inclusion-Exclusion Principle to eliminate subsets which violate one or more of the restrictions. – N. F. Taussig Apr 04 '22 at 09:25

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