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Wmwhite9
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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?!?
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)!} ## .Wmwhite9 said:How many ways can you arrange 52 things into 4 groups BUT the groups do not have to be the same size?!?
There are 52!/(13!)^4 ways to arrange 52 things into 4 groups.
Yes, the 52 things can be divided into groups of unequal sizes. However, the total number of arrangements will depend on the specific sizes of the groups.
Yes, the same thing can be in more than one group. This is known as a combination, where order does not matter.
There are 52!/(13!)^4 * 4! ways to arrange 52 things into 4 groups of equal size. This takes into account the different ways the groups themselves can be arranged.
Yes, the groups can be arranged in a specific order. This will result in a different total number of arrangements, as the order of the groups will also be considered in the calculation.