$\begingroup$How many different sums of the dots can one obtain if three ordinary dice are thrown at the same time? (an ordinary standard die is a regular cube with its six sides numbered with dots from 1 to 6)?
How should I approach the problem? I have tried to calculate it using combinatorics: ${6 \choose 3}=20$. In this count, we have counted the sum $8$ twice, for example. How can I find all the sums we have counted twice or more?
$\endgroup$1 Answer
$\begingroup$You can of course throw any integer from $3$ $(1,1,1)$ to $18$ $(6,6,6)$, so that gives $18-3 +1 =16$ options.
If you want to count number of the options for each sum, that number for sum $k$ is the coefficient of $x^k$ in the expansion of
$$(x + x^2+ \ldots + x^6)^3$$ which can be found by using standard series expansions.
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