Let $m$ white balls and $n$ black balls. We arrange them randomly in a row.
Let $k_1,\ldots, k_j \in\mathbb{N}$ such that $k_1 + \cdots + k_j = m$.
What's the probability the balls arrangement is:
$k_1$ white balls, 1 black ball, $k_2$ white balls, 1 black ball ...
The answer is $\frac{m!n!}{(n+m)!}$. The denominator is easy to understand, that's all the possibilities. How to understand the numerator?
$\endgroup$ 31 Answer
$\begingroup$Hint: The answer is $\frac{1}{\binom{m+n}{n}}$.
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