I've looked into some other relevant questions but I think mine is a bit more general. I'm trying to find the number of all the possible scenarios when splitting 10 people to groups of any number.
From what I remember from my probability courses, I can find the number of teams when they are exactly 2. They can either be groups of 8/1 , 7/2 , 6/3 etc. For any of those I just have to find the combinations of 8 people ,7, 6 etc and add them all to get all the possible 2 team combinations of any number of members.
Now the same thing must be done for 3 team and then added to the number we found for 2 teams. Here's where I get confused. Then we go on for 4 teams , 5 teams all the way up to 10 teams of 1. Finally of course we add the numbers of all these cases .
Can I find the total number more easily?
You are asking for the number of ways to divide $10$ people into groups, which is the same as the number of equivalence classes on a set of $10$ elements. This is the Bell number $B_{10}$.
EDIT This is in response to the comment by mm8511. I tried to make it a comment, but I couldn't type the formula correctly with seeing a preview.
There is a triangle scheme for computing the Bell numbers. There is also an exact formula in terms of a finite sum $$ B_n= \left\lceil\frac{1}{e}\sum_{k=1}^{2n}\frac{k^n}{n!}\right\rceil$$ This formula is problem $51$ in "The Art of Mathematics" by Bela Bollobas, where it is attributed to L. Comtet.