Configuration probability of partitioned objects

[rabbed]

With N objects, if I arrange each without replacement into K distinct partitions in which different object orders should not be accounted for:

- For distinct objects I get a total number of Wtot = K^N ways to partition them, and a specific distribution with N1 objects in partition 1, N2 objects in partition 2 etc. can be accomplished in W = N!/(N1!*N2!*...*NK) ways.

- For identical objects I get a total number of Wtot = (N+K-1)!/(N!*(K-1)!) ways to partition them, and a specific distribution with N1 objects in partition 1, N2 objects in partition 2 etc. can be accomplished in W = 1 way.

So the probabilities of a specific configuration in the two cases should be:

Pdistinct = N!/(N1!*N2!*...*NK) / K^N
Pidentical = 1 / ( (N+K-1)!/(N!*(K-1)!) )

Is this correct?

 

[tnich]

That does look correct.
If you are interested in the number of ways to allocate $n$ identical objects to $k$ distinct partitions, consider the coefficients of the polynomial $P_k(x) = \sum_{n=0}^{\infty} a_{k,n}x^n = (1-x)^{-k}$.