Permutations with Repetition and Repeated Elements

  • Thread starter Big-Daddy
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The theory says that if you have x objects containing a repeats of one element and b repeats of another, Np(without repetition)=x!/(a!b!).

If you have x objects and repetitions are allowed, Np(with repetition)=xx, correct?

Combining these, if we have x objects containing a repeats of one element and b repeats of another, and repetitions are allowed, Np(with repetition)=xx/(a!b!). Is this right?
 

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