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

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The discussion centers on the combinatorial problem of arranging 52 distinguishable objects into 4 distinguishable groups, where group sizes can vary and groups can be empty. The solution for this scenario is calculated as \(4^{52}\). Additionally, when the objects are indistinguishable, the problem translates to finding the number of non-negative integer solutions to the equation \(x_1+x_2+x_3+x_4=52\), which can be solved using the formula \(\binom{n+k-1}{k-1}\), where \(C\) denotes the binomial coefficient.

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How many ways can you arrange 52 things into 4 groups BUT the groups do not have to be the same size?!?
 
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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}##.
 
Wmwhite9 said:
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)!} ## .
 

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