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?
 A: 
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.
A: 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 $\frac{72}{13\choose 3}$
A: 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.)
