Clarification needed to understand elementary combinatorics problem 10 objects are randomly distributed among 3 boxes. What is the probability to have  6 objects in one of the boxes, 3 in another one and a single object in the remaining third box. 
My solution is 
For the first box we have $10 \choose 6$, since 6 objects are already placed we are left with 4 remaining, so for the second box we have $4 \choose 3$ and the remaining single object is placed in the last unoccupied box. Therefore, the total number of possibilities for creating the desired distribution of object among the 3 boxes is :
$$
{10 \choose 6} \times {4\choose 3} \times {1 \choose 1}
$$
To use the probability formula I need the total number of ways to distribute 10 objects among 3 boxes. 
Well, I cheated and looked in the answers section, the answer is $3^{10}$ and the probability is 
$$
P=\frac{{10 \choose 6} \times {4\choose 3} \times {1 \choose 1}}{3^{10}}
$$  
The question is how the $3^{10}$ can be explained. This looks like variation with repetition, but I cant really understand it.  
 A: $3^{10}$ is simply the total number of ways in which you can distribute to the three boxes, as for each object (out of ten) you have three potential boxes in which you can put it. For each object, there are $3$ potential boxes. We have $10$ objects. Thus by the multiplication rule, we have 
$$
3 \cdot 3 \cdot  3 \cdot 3 \cdot 3 \cdot 3 \cdot .... = 3^{10}
$$
ways in total of arranging the objects. 
In this question and more generally, you must divide your number of valid solutions (which you evidently found correctly) by your entire sample space to get your desired result.
A: To find the probability of a given random event occurring, you find a ratio $\frac{p}{q}$ where $p$ is the number of ways the event can occur and $q$ is the total number of ways the situation can occur. You correctly discovered $p$, but forgot to account for $q$. 
To see that $q = 3^{10}$, consider how we would go about placing the objects arbitrarily. We would start with the first object, and place it in one of three boxes, giving us 3 ways to do that. The second object has a similar choice, independent of where the first object went since we're making no restrictions on the situation. We repeat this for each ball. So we have 10 independent choices to make with 3 choices for each one, giving us $3 \cdot 3 \cdots 3 = 3^{10}$ ways.
