TITLE: What is the coefficient of $x^{18}$ in the expansion of $(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4}$?
QUESTION [5 upvotes]: How to approach this type of question in general? 

How to use binomial theorem?
How to use multinomial theorem?
Are there any other combinatorial arguments available to solve this type of question?

REPLY [14 votes]: We really seek the coefficient of $x^{14}$, factoring out an $x$ from each term in the generating function. Then observe that:
$(1 + x + x^{2} + x^{3} + x^{4} + x^{5}) = \frac{1-x^{6}}{1-x}$
Now raise this to the fourth to get: $f(x) = \left(\frac{1-x^{6}}{1-x}\right)^{4}$.
We have the identities:
$$(1-x^{m})^{n} = \sum_{i=0}^{n} \binom{n}{i} (-1)^{i} x^{mi}$$
And:
$$\frac{1}{(1-x)^{n}} = \sum_{i=0}^{\infty} \binom{i + n - 1}{i} x^{i}$$
So we expand out the numerator and denominator, picking terms of $x^{14}$. Note that we are multiplying the numerator expansion by the denominator expansion.
$$\binom{14 + 4 - 1}{14}x^{14} - \binom{4}{1} \binom{8 + 4 - 1}{8} x^{14} + \binom{4}{2} \binom{2 + 4 - 1}{2} x^{14}$$