Five balls into four boxes: no box left empty

Question

Five distinguishable balls are each dropped independently and uniformly at random into one of four distinguishable boxes. What is the probability that every box ends up with at least one ball?

Accepted Answer

There are 4^5 = 1024 equally likely outcomes. With four boxes, inclusion-exclusion now runs three correction terms. Let Ai be the event box i is empty: - one box empty: \binom{4}{1}\,3^5 = 4 	imes 243 = 972, - two boxes empty: \binom{4}{2}\,2^5 = 6 	imes 32 = 192, - three boxes empty: \binom{4}{3}\,1^5 = 4 	imes 1 = 4. Alternating signs give the bad count 972 - 192 + 4 = 784. Favorable (no empty box) = 1024 - 784 = 240, so P(	ext{no empty box}) = \frac{240}{1024} = \frac{15}{64} \approx 0.2344. The 240 is the number of onto assignments from 5 balls to 4 boxes. Takeaway. With four boxes th…

Five balls into four boxes: no box left empty

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