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!
 A: First, count the number of integer solutions to the equation $$x_1+\ldots+x_6=5$$ with $0\le x_i\le 2$, then calculate the number of integer solutions to that equation with only $0\le x_i$, and finally divide these two.
How can one do that? This is well explained in this answer.
(By the way, the correct answer is exactly 50%)
A: Without the restriction of no more than 2 balls per slot, there are exactly as many ways as there are functions
$$
f : [5] \to [6],
$$
and there are exactly $6^5 = 7776$ such functions.
With the restriction that we don't want more than 2 balls per any slot, I would probably break it into the cases of 1) at most one ball per slot, or 2) at least one slot has 2 balls.
The first case contains exactly $6!=720$ ways (choose the left-over slot, then put the 5 balls in the remaining 5 slots).
For the second case, we have to choose which slots get exactly 2 balls: There could be either 1 or 2 such slots with exactly two balls.
Further subcases:
Exactly 1 slot gets 2 balls: Choose it, choose the two balls to put in, then distribute 3 balls among 4 slots. There are
$$
\begin{pmatrix}6 \\ 2\end{pmatrix}\begin{pmatrix} 5 \\ 2\end{pmatrix}\begin{pmatrix} 4 \\ 1\end{pmatrix}3! = 3600
$$
ways to do this. 
Exactly 2 slots get 2 balls: Choose them, their balls, then put remaining ball in one of 4 remaining slots, for a total of 
$$\begin{pmatrix}6 \\ 2\end{pmatrix} \cdot 2! \cdot \begin{pmatrix}4 \\ 2\end{pmatrix} \begin{pmatrix}4 \\ 1\end{pmatrix}= 720$$ ways.
So the probability I found is $5040/7775 \approx 65$%. Clearly my answer doesn't agree with the one given by Hanz, perhaps someone will be able to point out the error I've made. I would like to say this approach is slightly more intuitive, although evidently more prone to errors!
A: 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.
