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probability of number of balls in the bucket

However, when it comes to labelled balls, an e.g.f. is waiting somewhere to be written.

Assume that the capacity of a bucket is 23. Then the e.g.f. for one bucket is:

$$ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{23}}{23!} $$

for ten distinct buckets one will have :

$$ (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{23}}{23!})^{10} $$

We are now interested in the coefficient of

$$\frac {x^{180}} {180!}$$, which agrees with previous calculus.

316493908895554453369790961313619572009615365012907112082570161990794872355520000505402092717575252190566769024926332221764288309486803949091940966173304490507414724761824000000000

The probability not to spill something is then 0.3164934.