number of answers of $2w+3x+5y+7z=n $ I know that we can use either Generating Functions or The Principle of Inclusion and Exclusion to answer this question and mostly the purpose of using generating functions is to make the calculations easier, so when we use generating functions we get:
$f(x) = \frac{1}{1-x^2}\frac{1}{1-x^4}\frac{1}{1-x^5}\frac{1}{1-x^7} $
but this does not make the calculations easier for this equation.
so the question is is there an easier way to actually count the answers? 
 A: I completely agree with you that if the only thing you're able to do is write down a generating function, then there is essentially no progress. If, on the other hand, you can analyze the generating function enough to extract its coefficients or, as is more often the case, approximate them, that is a different story.
Your generating function $f(x)=\frac1{(1-x^2)(1-x^3)(1-x^5)(1-x^7)}$ is a rational function. Its singularities are all on the unit circle at roots of unity ${\mathrm e}^{2\pi i j/k}$, according to the weights $k=2,3,5,7$. The singularity at 1 has multiplicity 4; the other thirteen singularities all have multiplicity 1. So, the partial fractions decomposition of the generating function consists of seventeen terms of the form $\frac\alpha{(1-x/\omega)^{1+\ell}}$, where $\alpha$ is a constant, $\omega$ is one of the roots of unity, and $\ell$ is a natural number less than the multiplicity of $\omega$. The coefficient of $x^n$ in such a term is $\alpha\binom{n+\ell}\ell/\omega^n$. Out of all the coefficients in the partial fraction decomposition, the contribution with biggest order of magnitude comes from the term with the highest multiplicity. In this case, it is the singularity at 1 with multiplicity 4 ($\ell=3$). So,
\begin{eqnarray*}
[x^n]f(x)&\approx&\alpha\binom{n+3}3\\
&=&\frac{\alpha n^3}{3!}+o(n^3).
\end{eqnarray*}
To compute the constant $\alpha$, you can multiply the generating function and its partial fraction decomposition by $(1-x)^4$ and evaluate the limit at 1. All the terms on the partial fraction side (except for the term with $\alpha$) go to zero, and with l'Hopital's rule or finite geometric series or whatever else, $\lim_{x\to1}(1-x)^4f(x)=\frac1{2\cdot3\cdot5\cdot7}$, resulting in  $\alpha=\frac1{2\cdot3\cdot5\cdot7}$. The approximation for the coefficients of the generating function is then
$$[x^n]f(x)=\frac{n^3}{2\cdot3\cdot5\cdot7\cdot3!}+o(n).$$
Actually, this entire computation is an instance of Schur's theorem in combinatorics (as opposed to the slew of   theorems named after him in other disciplines). Also, there's a nice write-up of your exact question in Wilf's generatingfunctionology book, section 3.15 page 96.
