Bunuel wrote:
In how many ways 50 different objects can be divided in 5 sets three of them having 12 objects each and two of them having 7 objects each ?
A. \(\frac{50!}{(12!)^3(7!)^23!}\)
B. \(\frac{50!}{(12!)^3(7!)3!}\)
C. \(\frac{50!}{(12!)(7!)^23!}\)
D. \(\frac{50!}{(12!)(7!)3!}\)
E. \(\frac{50!}{(12!)(7!)}\)
First, we can randomly distribute the 50 objects so start with 50! allocations (I prefer to imagine we're randomizing 50 numbers then splitting them into groups of the first 12, the next 12, the next 12, the next 7, then the final 7). If we focus on one of the sets, let's say one of the sets with 7 objects, this method of randomizing will create 7! of the same set for that particular set. Similarly, for each set of 12 it will create 12! of that same set, but in a different order. Therefore we should divide by 12! three times and 7! twice to
eliminate duplication within groups.
Finally, if we called the groups of 12's ABC and groups of 7's DE, we can see ABC is the same as ACB or BAC. There are 3! ways to get the same groups by swapping ABC's. Similarly, DE can swap but DE and ED are the same groups. So again divide by 3! then 2 to
eliminate duplication among groups.
Therefore I'd think the answer would be A but divided by another 2. Please correct me if I'm wrong!
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