Two men, each placing one ball in five out of ten boxes There are 10 boxes and each box can hold any number of balls. A man having 5 balls randomly puts one ball in each of the arbitrary chosen five boxes. Then another man having five balls again puts one ball in each of the arbitrary chosen five boxes. The probability that there are ball(s) in at least 8 boxes. Assuming all boxes and balls are identical. Both men can choose any arbitrary box.
My attempt:
I was really not able to pull through the solution. What I thought was to let the first man choose five boxes by $^{10}C_5$, and put the balls in those 5 boxes each. There is $1$ way to do that.
Now the second man comes and chooses his 5 boxes and put each ball in those 5 boxes. It can be done in $^{10}C_5 * 1$
Now these are the following ways the boxes can be filled:
$$1 1 1 1 1 1 2200$$
$$1111111120$$
$$1111111111$$
Now I am not able to understand how to apply combinatorics and find the number of ways these conditions can happen. Please correct me if I was wrong in my attempt at the solution.

Each of the three conditions you present represents a different way to fill the ten boxes with a different amount of balls so that at least 8 boxes have one ball. 
In the first case, you have two boxes with two balls, two boxes with zero balls, and six boxes with one ball. If you labelled the ten boxes, you are then looking for the amount of ways to choose 2 boxes to have two balls, 2 boxes to have zero balls, and 6 boxes to have one ball. Solving for this case gives:
\begin{align*}
{10\choose2}{10 - 2 \choose 2}{10 - (2 + 2) \choose 6}
= {10\choose2}{8\choose2}{6\choose6} =
\frac{10!}{2! * 2! * 6!}
\end{align*}
Repeat this for your remaining two cases, and you will have all the ways to arrange the balls in the ten boxes so that you have at least 8 boxes with 1 ball.
