Number of distinct sums possible for this card question?

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?

Answer 1

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!

Answer 2

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:

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