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How many ways are there to pick a collection of 12 coins from piles of pennies, nickels, dimes, and quarters?

a) Assuming that each pile has at least 12 or more coins.

For this one, I got CR(4,12) = C(4+11,12) = C(15,12) = 15! divided by(12!) and 3! = 455


and


b) Assuming that each pile has at least 12 or more coins and the pick must consist of at least one quarter coin or at least one dime.

For this one, I did 12 choose 4 minus 10 choose 2 to get 450.

Are those right? Any help would be appreciated! Thanks!

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  • $\begingroup$ Can you please clarify? How many pennies, nickels, dimes and quarters are there in the pile, and what is the size of the pile? Are there any restrictions on what the 12 coins should be? $\endgroup$ – Noy Soffer Jun 9 '16 at 20:12
  • $\begingroup$ It's not specified. It's supposed to be random. No restrictions. We're just looking at the fact that there are 12 coins with 4 different piles. $\endgroup$ – adhamncheese Jun 9 '16 at 20:21
  • $\begingroup$ How many coins are in each pile and what types of coins are in the piles? $\endgroup$ – Noy Soffer Jun 9 '16 at 20:22
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The answer to a) is correct.

For b), I do not understand the logic of the calculation. The way I would do it is to observe that the count is the answer to a), minus the number of bad choices, that is, choices with no dime and no quarter. By similar reasoning (Stars and Bars) as one you used for a), the number of bad choices is $\binom{13}{12}$. Or else one can count the bads without theory: We will use $0$ pennies, or $1$, or $2$, and so on up to $12$, a total of $13$ possibilities.

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  • $\begingroup$ I was a little confused on part B. Although this is a perfect and simple explanation. It makes sense now. Thank you very much! $\endgroup$ – adhamncheese Jun 9 '16 at 20:44
  • $\begingroup$ Although I feel like we're missing something though here. You explicitly used "choices with no dime AND no quarter". The question says OR. At least 1 quarter OR at least 1 dime. $\endgroup$ – adhamncheese Jun 9 '16 at 20:50
  • $\begingroup$ The good choices have at least one quarter or at least one dime. So the bad choices have neither. There are other less pleasant ways of counting the good choices directly: (i) at least one dime, no quarter; (ii) at least one quarter, no dime; (iii) at least one of each. For example, for at least one of each we grab a dime and a quarter, and need to choose $10$ coins. There are $\binom{13}{10}$ ways to do this. Counting (i) and (ii) is similar. $\endgroup$ – André Nicolas Jun 9 '16 at 20:56
  • $\begingroup$ See, I guess what I am not understanding is if the good choices have at least 1 qtr or at least 1 dime, how would the bad choices have neither? Shouldn't it be at least 1 or at least the other that isn't with the bad choices? $\endgroup$ – adhamncheese Jun 9 '16 at 21:01
  • $\begingroup$ You will be happy if there is at least one dime or at least one quarter, or at least one of each. So when will you be unhappy? When dimes and quarters are entirely missing. $\endgroup$ – André Nicolas Jun 9 '16 at 21:04

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