How many three digit positive integers are there, the sum of whose digits is a perfect cube? How many three digit positive integers are there, the sum of whose digits is a perfect cube?
I already have an answer. My answer is 24, but when I inspected the answer given, it gave 38. I do not know where I went wrong. 
Here is what I did. 
Case 1: sum is 1, only 1 exists 100, $1+0+0=1$
Case 2: sum is 8, 
a.) 800, $8+0+0=8$
b.) by stars and bars,$ a+b+c=8,$ ${n-1 \choose k-1}$ hence ${7 \choose 2}=\frac {(7)(6)(5)!}{(2)!(5)!}=21$ 
Case 3: sum is 27, only 1 exists 999, $9+9+9=27.$
Therefore, adding up all the 3-digits integers I gathered from cases 1 to 3, I get $ 1+22+1=24$
Can anyone spot where I went wrong? Or do you all agree with me?

For sum $8$, you must start with a "star", and after that have 7 stars and 2 bars, so ${7+2\choose 2}=36$.
