the coefficient of $t^{18}$ in $t^6(1+t+\cdots+t^5)^6 $ equals the coefficient of $t^{12}$ in $(1+t+\cdots+t^5)^6 $.

Question

Let $X$ be the total from rolling 6 fair dice, and let $X_1,\ldots,X_6$ be the individual rolls. What is $P(X=18)$? Then the PGF of $X$ is $$E(t^X) = E(t^{X_1} \cdots t^{X_6}) = \frac{t^6}{6^6}(1+t+ \cdots+t^5)^6$$ So the number of ways to get a sum of $18$ is the coefficient of $t^{18}$ in $t^6(1+t+\cdot\cdot\cdot+t^5)^6$, which is the coefficient of $t^{12}$ in $(1+t+\cdot\cdot\cdot+t^5)^6 $.

I don't understand why the coefficient of $t^{18}$ in $t^6(1+t+\cdot\cdot\cdot+t^5)^6$ is the same as the coefficient of $t^{12}$ in $(1+t+\cdot\cdot\cdot+t^5)^6 $?

Answer 1

the first one, you'll note, has $t^6$ glommed onto the beginning of its generating function; remove that, and you go from the coefficient of $t^{18}$ to that of $t^{18-6}=t^{12}$.

Answer 2

When you expand $(1+t+\dots+t^5)^6$, you get some polynomial of degree $30$, which we can write as $\sum_{n=0}^{30} a_n t^n$ for some $\{ a_n \}_{n=0}^{30}$. When you multiply it by $t^6$ you get a polynomial of degree $36$, specifically $\sum_{n=0}^{30} a_n t^{n+6}$. The coefficient of $t^{18}$ here is $a_m$ where $m+6=18$, that is, $m=12$. So the coefficient of $t^{12}$ in the first one is the coefficient of $t^{18}$ in the second one.

Answer 3

The number of ways to get $n$ in one throw of the dice is the coefficient of $n$ in $t+t^2+t^3+t^4+t^5+t^6$, in other words it is one for $n=1,2,3,4,5,6$. The probability to get, say, seven, in two throws of the dice is the value of the $t^7$ term in $(t+t^2+t^3+t^4+t^5+t^6)^2$. This is just a "coincidence" in that the number of ways to combine works out to be the same as the coefficient $t^7$. This is quite standard and explained in any textbook.

Answer 4

$$(1+t+\cdots+t^5)^6=1+\dots+at^{12}+\dots+t^{30}$$ $$t^6(1+t+\cdots+t^5)^6=t^6(1+\dots+at^{12}+\dots+t^{30})=t^6+\dots+at^{12+6}+\dots+t^{36}$$