Suppose I have a large set of elements of size $K = 40$, and I would like to assign elements to two smaller groups of size $n=25$ and $m=15$. In this case, what are the total number of possible permutations I can have? Would it be $40 \choose 25$ or would it be $\frac{40!}{25!15!}$?
2026-03-08 08:37:09.1772959029
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If I have a large set of elements of size $K$, what are the possible number of ways I can uniquely distribute them to two sets of size $n$ and $m$?
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I think you need to make sure you understand the concept of choosing~ 1)by definition ${m\choose n} =\frac{m!}{n!\times(m-n)!}$. 2)What does it mean by "choose"? If I have 5 people and I need to choose 2 out of 5 people, then I can take 2 people out, do I can care how the rest of the people are arranged? NO~ cuz I only need to choose 2 two people. That is why you need to have 40! divided by 15! because you are choosing 25 people. p.s. since the order doesn't matter you also need to divide 40! by 25!. So the answer to your question is that you are right either way~
Both answer are correct: ${40 \choose 25} = \frac{40!}{25!15!}$.