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Let consider an urn model containing $n$ balls such that $m_g$ balls are green, $m_r$ are red and $m_b$ are blue and we have $n = m_g + m_r +m_b$.

Especially this means that the $m_g$ green balls are indistinguishable with respect to each other (and the same game for red and blue ones)

We draw $l \le n$ times without placing the balls back.

My question how many posibilities there exist for different $l$-draws taking the order and the indistinguishability of balls of same color into account.

More generally: the urn contains $m_1, m_2, ..., m_k$ balls with $\sum _i ^k m_i =n$ such that the $m_j$ are undistiguishable and one make $l$ draws without placing the balls back.

What is the cardinality of the resulting probability space?

Could anybody explain if there exist an intuitive way to describe and the count all cases?

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  • $\begingroup$ Are you saying the order of the balls matters? That is, GBB is considered different from BBG? $\endgroup$ – Mike Earnest Jan 28 at 18:11
  • $\begingroup$ @MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right? $\endgroup$ – KarlPeter Jan 28 at 18:13
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The obvious answer is $$ \sum_{\substack{a_1+a_2+\dots+a_k=\ell\\0\le a_i\le m_i}}\binom{\ell}{a_1,a_2,\dots,a_k} $$ but this is an unwieldy summation.

A less obvious answer: Let $E_m(x)=\sum_{i=0}^m \frac{x^i}{i!}$ be the partial exponential series. The size of your sample space is $$ \ell![x^\ell]\prod_{j=1}^k E_{m_j}(x) $$ Here, $[x^\ell] f(x)$ denotes the coefficient of $x^\ell$ in the power series $f(x)$. This at least lets you compute the size much quicker, since polynomials can be multiplied quickly using the Fast Fourier Transform.

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