How many ways can you arrange 52 things into 4 groups BUT th

[Wmwhite9]

How many ways can you arrange 52 things into 4 groups BUT the groups do not have to be the same size?!?

 

[Khashishi]

If the things are distinguishable and the groups are distinguishable and a group can be zero size and the order within a group doesn't matter, $4^{52}$.

 

[WWGD]

How many ways can you arrange 52 things into 4 groups BUT the groups do not have to be the same size?!?

If I understood correctly, this is the problem of balls in boxes, where you want to put 52 objects in 4 boxes. If the things are indistinguishable, then the answer is the solution to $x_1+x_2+x_3+x_4=52$ , which is equal to $(n+k-1)C(k-1)$ , where $C$ stands for "choose", as in " x choose k":= $\frac {x!}{k!(x-k)!}$ .