# How come the probability here is not $1/({_{13}C_3})$?

Here's my question.

I have $6$ gold coins, $4$ silver coins and $3$ bronze coins in my pocket. I take out three coins at random. What is the probability that they are all of different material?

I had thought that the answer was $\frac{1}{13 \choose 3}$. However, the answer turned out to be $\frac{72}{13 \choose 3}$. Why?

Isn't there only one combination of 3 coins that allows for the coins to be different?

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The coins can be thought of as being distinct. There are 6 ways to choose a gold coin, 4 ways to choose a silver coin, and 3 ways to choose a bronze coin. So, by the multiplication principle... –  David Mitra May 23 '12 at 21:27
What you have is the probability for drawing either three coins out of the 13. You need to multiply with the number of draws that result in exactly one gold, one silver and one bronze. –  M.B. May 23 '12 at 21:30
Ah thanks, but doesn't ${13}\choose{3}$ return the total number of combinations regardless of whether the coins are distinct? –  John Hoffman May 23 '12 at 21:38
What do you mean? 13C3 is the total number of ways you can draw 3 from 13 without replacement. –  M.B. May 23 '12 at 21:41
@Michael Hardy Why did you change $13C_3$ back to $_{13}C_3$? –  user17762 May 23 '12 at 22:52

Isn't there only one combination of 3 coins that allows for the coins to be different?

Nope. For the coins to be different, one must choose 1 gold coin (6 possibilities), 1 silver coin (4 possibilities) and 1 bronze coin (3 possibilities), for a total of 6.4.3=72 possibilities. This is the numerator of the correct answer.

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Just to understand better, imagine there were $2$ gold coins, $2$ silver coins and $1$ bronze coin. Then if you label them as $G_1 G_2 S_1 S_2 B_1$, The combination of three distinct coins would be

$\{G_1 S_1 B_1\}, \{G_1 S_2 B_1\}, \{G_2 S_1 B_1\}, \{G_2 S_2 B_1\}$

Essentially which is $2 \times 2 \times 1 = 4$ distinct combination of choice.

The same way you get a total of $72$ unique combinations, and thus you are getting the answer expected as $\frac{72}{13\choose 3}$

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The coins can be thought of as being distinct. There are 6 ways to choose a gold coin, 4 ways to choose a silver coin, and 3 ways to choose a bronze coin. So, by the multiplication principle, there are $6\cdot 4\cdot 3$ ways to choose three coins such that all are of a different material. Note this counts the number of unordered selections. (But, you can keep track of things by imagining that you choose a gold coin first, then a silver coin, and finally a bronze coin.)

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