The problem is "You are making 100 cookies. How many chips $n$ do you need to put into the batter to have at least 90% probability that every cookie has at least one chip?"
I tried a stars and bars combinatorics approach to this as follows:
p(every cookie has at least one chip) = $\dfrac{\# permutations \;with \geq 1 \; chip \; per \; cookie}{\# total \; permutations}$
Given 99 'bars' (cookie separators) amongst $n$ 'stars' (chips) there are $\binom{n+99}{99}$ possible permutations of chips across the cookies (permutations not combinations because we can imagine the bars dividing chips between 100 cookies numbered 1-100)
We can tweak the stars and bars approach for the numerator, i.e. consider 99 bars/cookie separators which can be allocated (without repetition) to the $n-1$ spaces between stars/chips. This yields $\binom{n-1}{99}$ permutations where every cookie has at least one chip, as we have left out the edges (hence $n-1$) and ensured that bars cannot be adjacent (which would result in 0-chip cookies) — there is at least one chip between each bar.
Giving an answer which is the solution to $\dfrac{\binom{n-1}{99}}{\binom{n+99}{99}} = 0.9$
This would need to be calculated numerically of course (as does the exact solution in the book), but plugging in the correct answer of $n=683$ yields a tiny probability, $<10^{-6}$.
Clearly I have gone wrong somewhere, so my question is, where?