Expanding a probability generating function for the total score of rolling a die r times An unbiased six-faced die is rolled r times. The probability generating function for the total score is
$[\frac{t(1-t^6)}{6(1-t)}]^r$
Hence show that the probability of the total score being (r+3) is
$\frac{1}{6}^{r+1}r(r+1)(r+2)$
I appreciate any help anyone can provide and this is a statistics question in a Further Maths A-Level style paper for context on how they want the question to be solved. 
 A: Let $[t^n] f(t)$ denote the coefficient of $t^n$ in $f(t)$.  (I'm not sure how standard this notation is but it's very useful.)  There is a factor of $t^r$ in the generating function.  
$$[t^{r+3}] \left( t (1-t^6) \over 6(1-t) \right)^r = [t^3] \left( 1-t^6 \over 6(1-t) \right)^r $$
and you can pull out a factor of $(1/6)^r$ as well to get
$$ [t^3] \left( {1-t^6 \over 1-t} \right)^r $$
At this point there are two possible methods.
(1) Note that
$$\left( {1-t^6 \over 1-t} \right)^r = (1 + t + t^2 + t^3 + t^4 + t^5)^r$$
and think about how multiplication of polynomials works to get the coefficient of $t^3$ - you can get a factor of $t^3$ from one factor of $t^3$ and the rest 1, or a $t^2$ and a $t$, or three copies of $t$.  (This isn't all that different from solving the problem without generating functions.)
(2) Take the third derivative and evaluate at zero.  Since you have an $r$ in the exponent this might be a bit tricky - I'd try logarithmic differentiation.  I have not actually done this.
A: Picking up from where Michael Lugo left off, we want to compute $$[t^3]{(1-t^6)^r \over (1-t)^r}.$$ Expand the numerator via the binomial theorem and drop any terms that involve a power of $t$ greater than $3$: $$[t^3]{(1-t^6)^r \over (1-t)^r} = [t^3]{1\over (1-t)^r}.$$ Now use the identity $${1\over(1-s)^{k+1}} = \sum_{n=0}^\infty \binom{n+k}k s^n$$ to get $$[t^3]{1\over (1-t)^r} = \binom{r+2}{r-1} = \binom{r+2}3 = {r(r+1)(r+2)\over6}.$$ Multiply this by the factor of $1/6^r$ that was pulled out earlier to get the final result.
