In how many ways can we put 10 distinguishable people into 2 groups so that no group is empty? Order does not matter in a group and groups are not named.
So I've done some research and people seem to say that this is a Stirling problem of the second kind.
This is the same as the number of ways to distribute n distinct balls into k identical boxes so that no box is empty = $S(n, k)$
In particular $S(n=10,k=2) = 511$
Is this correct? Also is there a more logical approach to this problem?
I think there that are $2^{10}$ ways to place $10$ distinguishable balls into $2$ boxes. However, since the boxes are indistinguishable, we are overcounting by $2!$. Therefore, $2^{10}/2 = 2^9$
Using inclusion-exclusion, we subtract the case where $1$ box has all the balls and the other has $0$. There is only $1$ way this can happen since the teams are indistinguishable and we don't care about order.
Therefore, $512-1=511$
Best Answer
That is correct. Stirling Number of the second kind or the Principle of Inclusion Exclusion is the right approach for such problems. But just consider this in your specific case and simpler cases like this can be solved without using Stirling Number of the second kind.
Every person has two choices of groups and so across $10$ people, there are $2^{10}$ choices. But given the groups are not distinguishable, the choices are $\frac{1}{2} \times 2^{10} = 512$.
Now there is only one way all of them form just one group and hence,
Number of ways to form two non-empty groups $ = 512 - 1 = 511$.