How many different sums can be formed excluding double counted sums The question asks to find the number of different sums (combinations) which can formed from coins valued:
$(5), (1), (0.50), (0.25), (0.10), (0.03), (0.02)$, and $(0.01)$; but avoid the double counted sums.
I have solved it as:
Assuming no double counted sum (i.e $0.01 + 0.02 = 0.03$; to account for this I've subtracted $31$ below), $$({2^8} - 1) - ({2^5} - 1) = 255 - 31 = 224$$
Have I got it right?
 A: To clarify, your question asks for the number of possible sums you can make from coins of this value using each coin at most once? In that case, I agree with most of your work. The only possible "redundancy" with the coins is having $.01+.02$ instead of $.03$, since for all coins $c$ besides $.03$, the sum of the coins less than $c$ is still less than $c$.
Assuming you don't want the empty sum (though I'd argue that the sum of $0$ coins is a sum), there are $2^8-1$ combinations. The number of double-counted combinations is the number of combinations that use $.03$ and not $.02$ or $.01$, since there is an equivalent combination obtained by keeping the other coins the same and swapping $.03$ for $.01+.02$. What I don't agree with is the $2^5-1$. Even if you want to exclude the empty sum, it shouldn't matter here, because you'll still have $.03$ or $.01+.02$, so it's not counted here. In fact, it should be clear that the empty sum is counted only once. Hence, there are $2^5$ combinations that are double-counted, so you subtract them from $2^8$.
