Number of distinct sums possible for this card question? 
There are $8$ cards with number $10$ on them, $5$ cards with number
  $100$ on them and $2$ cards with number $500$ on them. How many
  distinct sums are possible using from $1$ to all of the $15$ cards?

This first time I looked at this question it seemed quite simple, but the more I work on it, the trickier it gets. I am beginner, who is trying to master combinatorics.
How do I approach this question? 
Initially I came up with an answer of $8\times5\times2=80$, but I definitely feel that there is something more to it. 
Where am I going wrong? 
 A: We can use the fact that $100\times5=500$.
Instead of considering $2$ cards numbered $500$, we could consider $2\times500=1000$ as $10\times100=1000$, i.e., we consider $10$ cards which are numbered $100$.
So, the total cards are : $8$ cards numbered $10$ and $15$ cards numbered $100$.
Now the task at hand is to compute the number of possible sums we could get using at least $1$ card.
There are ($8,7,6...,0$) i.e., 9 possibilities for cards which are numbered $10$.
And there are ($15,14,...,0$) i.e., 16 possibilities for cards which are numbered $100$.
That is, $$9\times16 = 144$$ possibilities.
Again, considering the fact that question asks to consider that at least $1$ for the sum, that is, we have to exclude the case where we don't pick a card at all, the answer is $144-1=143$ distinct sums possible!
A: There are 9 ways of choosing 0 or more 10s i.e [0,8]
There are 6 ways of choosing 0 or more 100s i.e [0,6]
There are 3 ways of choosing 0 or more 500s i.e [0,2]
So total no. of sums possible = 9x6x3.
But we don't want 0 sum and duplicate sums.
We can see that duplicate sums are only possible in the below two cases:

*

*5 100s and 0 500s = 0 100s and 1 500 = 500 (we
get 9 duplicate)

*5 100s and 1 500 = 0 100s and 2
500s = 1000 (we get 9 duplicate)

Final answer = 9x6x3 - 1 (for zero case) - 18 (for duplicates) = 143
