We want to determine the number of ways that 10 distinct pieces of candy can be placed into 2 identical bags with each bag getting at least 1 piece.  

As we know the answer is $S(10,2)$, where $S$ stands for Stirling number of the second kind.  

I thought that we can start by choosing two candies out 10 to put each one in a bag, so we will have that each bag is nonempty. Then, the remaining 8 candies will either be put in a single bag or put in the two bags (meaning each bag gets at least one candy out of 8) and that can be done in $S(8,2)$ ways.
And finally, I get $S(10,2)= \binom {10}{2} \left( S(8,1) + S(8,2) \right) $.