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I have 3 subsets: A, B and C, each have 100 elements. How can I calculate how many combinations I can have of A B and C for a total of 100 elements in a new subset.

JMP
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JR 19
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Feb 01 '23 at 11:00
  • Please edit to include your efforts. Also to clarify the question...as it stands, I am not sure what you are asking. Perhaps it would help to work it out explicitly for some number smaller than $100$. Maybe $3$ is good enough to clarify the situation. – lulu Feb 01 '23 at 11:03
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    Must you have at least one each from $A,B,C,$ or is it permissible to have, say, $A=0, B=0, C=100$ – true blue anil Feb 01 '23 at 11:22
  • I should clarify. One combinaiton would be 1 element A, 1 element B and 98 element C. Another combination would be 1 element A, 2 element B and 97 element C and so on. – JR 19 Feb 01 '23 at 11:30
  • @JR19 you should add your clarification to the question part , because the question and your actual purpose are different – Not a Salmon Fish Feb 01 '23 at 16:43

2 Answers2

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From your clarification, it seems clear that You are asking for combinations, thus order doesn't matter, and that there must be at least $1$ each from $A,B,C$

This is easily solved by a version of "stars and bars" which is explained here (you need not see the ADDED portion)

Using the formula $\dbinom{n-1}{k-1}$,

the answer is $\dbinom{100-1}{3-1} = \boxed{4851}$

true blue anil
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From what I understand the question seems rather simple, so maybe a clarification on what you asked is needed, but for now: let's assume you don't care about order and there's no limits on how many elements from A, B or C you need to get every time, then we can consider a general total pool of $300$ elements. Since there are $\tbinom {n}{k}$ ways to choose an (unordered) subset of $k$ elements from a fixed set of $n$ elements, you're answer would be $\tbinom {300}{100}=4,158,251,463,258,564,744,783,383,526,326,405,580,280,466,005,743,648,708,663,033,658,000,000,000,000,000,000$ hope this helps

Edit1: the clarification made on the comments takes out the cases $0,0,100$ and $0,100,0$ and $100,0,0$. It is still not clear what happens when we reach $1,98,1$, if order is counted or not and if something like $2,2,96$ is possible...