What is the coefficient of $x^{18}$ in the expansion of $(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4}$? 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?


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.
$$\binom{14 + 4 - 1}{14}x^{14} - \binom{4}{1} \binom{8 + 4 - 1}{8} x^{14} + \binom{4}{2} \binom{2 + 4 - 1}{2} x^{14}$$
