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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.

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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 $2^{5} = 32$.

That is, the dollar coins are really red-herrings; all they do is fill up the space. The question would have the exact same answer if it had been phrased as "In how many different ways can you select coins from among a single penny, a single nickel, a single dime, a single quarter, and a single half-dollar?"

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Decide for each small coin whether you select that or not. Then top up with dollars until you have selected 5 coins in total.

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.

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  • $\begingroup$ Your answer was good as well but the reason I chose Arturo's was because of the added explanation. :) $\endgroup$ Commented Dec 11, 2011 at 7:14

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