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What are the number of ways of dividing $n_1$ objects of type $1$, $n_2$ objects of type $2,\ldots,n_k$ objects of type $k$ into $2$ parts?

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up vote 2 down vote accepted

I think it is:$$\prod_{i=1}^{k}(n_i+1)$$ The reasoning is that for each i you can keep either ${0,1,2,....n_i}$ balls in the first part and the rest of the balls in the 2nd part so there are $(n_i+1)$ choices for each i. From this the result follows. This also includes the case when one part is empty.

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I think we should subtract one from the product to exclude the case when each i is assigned "0" right?? – user60576 Feb 3 '13 at 16:52

It's solved by first finding the $N$ "Permutations of Like Objects," and then applying the binomial coefficient.

Definition: The total number of permutations of all $n$ distinct objects for which $r_1$ are alike, $r_2$ are alike $\ldots, r_k$ are alike is

\begin{equation} N=\frac{n!}{r_1!r_2!... r_k!} \end{equation}

Then use the binomial coefficient to find the number of ways of selecting 2 objects from your $N$ permutations using

\begin{equation} N_{total}=\binom N 2= \frac{N!}{2!(N-2)!} \end{equation}

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