Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$ We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.
$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
 = 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$
Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.
 A: $$1+x+x^2+\cdots+x^5=\frac{1-x^{6}}{1-x}=(1-x^{6})(1-x)^{-1}$$
so
$$(1+x+x^2+\cdots+x^5)^4=\left(\frac{1-x^{6}}{1-x}\right)^4=(1-x^{6})^4(1-x)^{-4}=(1-4x^{6}+\cdots)(1+4x+10x^2+20x^3+35x^4+56x^5)$$
There is only one $x^4$ term, and you can see immediately that its coefficient is $35$.
A: This is of course :-) intended to find out the answer to:

You have a bag of $16$ (identical) frogs, and wish to distribute them to $4$ of your wizard friends. However each wizard only has enough pocket space to carry $5$ frogs. In how many ways can you distribute the frogs?

And the answer by inclusion-exclusion is
$\binom {16+4-1}{4-1} - \binom{4}{1}\binom {16-6+4-1}{4-1} + \binom{4}{2}\binom {16-12+4-1}{4-1} = \binom {19}{3} - 4\binom {13}{3} + 6\binom {7}{3} = 969 -1144+210 = 35$


Edit to add: As Doug M mentions in comments to the main question, this is also the number of ways to roll $20$ on four distinct dice, since choosing $4$ times from $\{0,1,2,3,4,5\}$ to make  a total of $16$ can be done the same number of ways as choosing $4$ times from $\{1,2,3,4,5,6\}$ to make $20$.
A: \begin{align} \left(1+x+x^2+x^3+x^4+x^5\right)^2\ \times \left(1+x+x^2+x^3+x^4+x^5\right)^2 \\
= \left(1+2x+3x^2+4x^3+5x^4+6x^5+\color{red}{5x^6}+\color{orange}{4x^7}+\color{green}{3x^8}+\color{blue}{2x^9}+\color{purple}{x^{10}}\right)\ \times \left(1+2x+3x^2+4x^3+5x^4+6x^5+\color{purple}{5x^6}+\color{blue}{4x^7}+\color{green}{3x^8}+\color{orange}{2x^9}+\color{red}{x^{10}}\right)\\
\\
\end{align}
So the coeffecient of $x^{16}$ is $\ \color{red}{5\times 1} + \color{orange}{4\times 2} + \color{green}{3\times 3} + \color{blue}{2\times 4} + \color{purple}{1\times 5} = 35.$
A: A monomial giving a contribution to degree 16 must be of the form
$$x^{a_1}\cdot x^{a_2}\cdot x^{a_3}\cdot x^{a_4}$$
with $a_i <6$ and $a_1+a_2+a_3+a_4=16$.
Suppose  $a_1 \geq a_2 \geq a_3 \geq a_4$, then you have only five possibilities: $(5,5,5,1)$, $(5,5,4,2)$, $(5,5,3,3)$, $(5,4,4,3)$ and $(4,4,4,4)$.
The coefficient of $x^{16}$ equals the cardinality of the set composed by these vectors and the ones obtained permuting their coordinates.
It is easy to prove that this cardinality is equal to:
$$\binom{4}{3}+2\binom{4}{2}+\binom{4}{2}+2\binom{4}{2}+1=35$$
