Doubt in finding number of integral solutions Problem : 
writing $5$ as a sum of at least $2$ positive integers. 
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $${7\choose3}= 35$$
but we can count the cases and say that answer must be $6$ :
$4 + 1 $
$3 + 2 $
$3 + 1 + 1$
$2 + 2 + 1 $
$2 + 1 + 1 + 1$
$1 + 1 + 1 + 1 + 1$
At which stage am I making a mistake ? Thanks

According to your generating function approach let's assume we want to find the number of compositions of $5$ having at least two parts.

Using generating functions we have to look for the coefficient of $x^5$ in
  \begin{align*}
&(x+x^2+x^3+\cdots)^2+(x+x^2+x^3+\cdots)^3+(x+x^2+x^3+\cdots)^4\\
&\qquad+(x+x^2+x^3+\cdots)^5\\
&\qquad=\frac{x^2}{(1-x)^2}+\frac{x^3}{(1-x)^3}+\frac{x^4}{(1-x)^4}+\frac{x^5}{(1-x)^5}\tag{1}\\
\end{align*}
  The coefficient of $x^5$ in each of the four terms denotes for $j=2,3,4,5$ the number of possibilities to write the number $5$ as sum of $j$ numbers $\geq 1$.

It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$.

We obtain for $j=2,3,4,5$
  \begin{align*}
[x^5]\frac{x^j}{(1-x)^j}&=[x^{5-j}]\frac{1}{(1-x)^j}\tag{2}\\
&=[x^{5-j}]\sum_{k=0}^\infty\binom{-j}{k}(-x)^k\tag{3}\\
&=[x^{5-j}]\sum_{k=0}^\infty\binom{k+j-1}{j-1}x^k\tag{4}\\
&=\binom{4}{j-1}\tag{5}
\end{align*}

Comment:


*

*In (2) we use the rule
$
[x^{p-q}]A(x)=[x^p]x^qA(x)
$

*In (3) we use the binomial series expansion

*In (4) we use the binomial identity
$
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
$

*In (5) we select the coefficient from $x^j$

We conclude from (1) and (5): The number of solutions is
  \begin{align*}
\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}&=4+6+4+1\\
&=15
\end{align*}
The corresponding $15$ tuples are
\begin{align*}
&14,23,32,41\\
&113,122,131,212,221,311\\
&1112,1121,1211,2111\\
&11111
\end{align*}

Hint: The generating function $(x+x^2+x^3+...)^2(1+x+x^2+...)^3$ is not appropriate since it counts e.g. $$1+2+2$$ more than once.
In fact $1+2+2$ is counted three times
\begin{align*}
(x+x^2+x^3+...)^2&(1+x+x^2+...)^3\\
1+2\quad\quad&\quad+2+0+0\\
1+2\quad\quad&\quad+0+2+0\\
1+2\quad\quad&\quad+0+0+2\\
\end{align*}
