# Simple Combinations with Repetition Question

In how many ways can we select five coins from a collection of 10 consisting of one penny, one nickel, one dime, one quarter, one half-dollar and 5 (IDENTICAL) Dollars ?

For my answer, I used the logic, how many dollars are there in the 5 we choose?

I added the case for 5 dollars, 4 dollars, 3 dollars and 2 dollars and 1 dollars and 0 dollars. $$C(5,5) + C(5,4) + C(5,3) + C(5,2) + C(5,1) + 1 = 32$$

which is the right answer ... but there has to be a shorter way, simpler way. I tried using the repetition formula that didn't pan out.

If you could introduce me to a shorter way with explanation I appreciate it.

This is equivalent to counting the number of subsets of the non-dollar coins, because you have exactly 5 (pairwise distinct) non-dollar coins, and you are trying to select $5$ coins. Each possible selection is completely determined by the subset of $\{\text{penny}, \text{nickel}, \text{dime}, \text{quarter}, \text{half-dollar}\}$ that it contains. So you just need to count the number of subsets of a set of 5 elements, which is $2^{5} = 32$.
The topping-up step does not involve any choice, so you have 5 choices to make, each with 2 options, giving $2^5=32$ combinations in all.