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How would I find the total number of combinations for any given group sets? For instance:

I have 5 fields,

Field A, B, C, D, and E.

A, D, and E need to be filled with 1 number. B and C need to be filled with 3 non repetitive numbers.

Field A and E can be 1 - 4, B can be 1 - 16, C can be 1 - 10, and D can be 1 - 11. Like this:

Category A B C D E

How many 1 3 3 1 1

Choices 4 16 10 11 4

So obviously the first possible outcome would be something like:

A - 1 B - 1,2,3 C - 1,2,3 D - 1 E - 1

Hopefully it makes sense and you can see that each columns choices are independent of each other, when this is in practice it will actually be unique names. If someone could help me out with a formula that would be fantastic. Thank you very much.

Michael
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  • I don't know if i'm doing this right, but i'm pretty sure I have this part down: A - has 4 options B - has 560 possible unique combinations C - has 120 possible unique combinations D - has 11 options E - has 4 options

    So then in order to get all possible combinations of each, do I just multiply all of that together to get 11,827,200?

     – Michael Oct 07 '15 at 00:47
  • Exactly right. Since the choices are independent of each other, they can be multiplied together. And the two that require 3 objects are both calculated as ${n \choose 3} = \frac{n!}{(n-3)!3!}$ – Paul Sinclair Oct 07 '15 at 02:07
  • You are correct about applying the multiplication principle for independent choices. The answer you get is $\binom{4}{1}\binom{16}{3}\binom{10}{3}\binom{11}{1}\binom{4}{1}$ and your numeric value looks correct. – Marconius Oct 07 '15 at 02:08

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