this post was submitted on 18 Jul 2023
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Unless I'm getting the math wrong myself, for any "pick 1" combination set like this we're dealing with just multiplying the combination sets together. Technically we're multiplying by the factorial of the sample size, but
1!=1.We're not picking any 10 from within the subset of 100; you cannot pick both ending 1 and ending 4 from companion A and then no ending at all for companion C. I'm assuming each individual sub-ending is mutually exclusive with the rest of its sample space. That difference of assumptions is what led to your 1.7x10^13^ combinations.