<s>[INST] What is the probability no slots contain more than two balls given I am trying to sort 5 balls into 6 slots? I am having a difficult time understanding how to approach this problem.  Suppose I have $6$ total slots and $5$ balls. Now, I assign the balls at random to the slots. What is the probability that no slot will contain more than two balls?
My approaches so far:
I recognized that $\binom{5+(6-1)} 5$ ($10$ choose $5$) represents the total combinations per the combinations with repetitions formula.
Next, I realized that maybe I can find the probability a slot has exactly $5$, $4$, $3$ balls in one spot, then take $1$ - the sum of those probabilities. 
The correct answer is around $80\%$ so I am way off with this approach. Do you guys have any idea? Thanks!
 [/INST] Five balls can be distributed in 6 slots following the GF
$(1+b+b^2+b^3+b^4+b^5)(1+b+b^2+b^3+b^4+b^5)...(1+b+b^2+b^3+b^4+b^5)$ (six factors)
The coefficient of $b^5$ is $252$ and represents the ways of placing 5 balls without restrictions in six slots. With the given restriction, the GF is
$(1+b+b^2)(1+b+b^2)...(1+b+b^2)$
The coefficient of $b^5$ is now $126$, that gives a probability of $50 \% $, as pointed above.
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