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

Actually you count all the possible permutations that gives $x^5$ in 
$\displaystyle \color{red}{\frac{x^2}{(1-x)^2}} \times
\color{blue}{\frac{1}{(1-x)^3}}$,
$\color{red}{(4+1)}+\color{blue}{(0+0+0)}$ counts 
$\color{red}{2} \times \color{blue}{1}$ possibilities.
$\color{red}{(3+2)}+\color{blue}{(0+0+0)}$ counts 
$\color{red}{2} \times \color{blue}{1}$ possibilities.
$\color{red}{(3+1)}+\color{blue}{(1+0+0)}$ counts 
$\color{red}{2} \times \color{blue}{3}$ possibilities.
$\color{red}{(2+2)}+\color{blue}{(1+0+0)}$ counts 
$\color{red}{1} \times \color{blue}{3}$ possibilities.
$\color{red}{(2+1)}+\color{blue}{(2+0+0)}$ counts 
$\color{red}{2} \times \color{blue}{3}$ possibilities.
$\color{red}{(2+1)}+\color{blue}{(1+1+0)}$ counts 
$\color{red}{2} \times \color{blue}{3}$ possibilities.
$\color{red}{(1+1)}+\color{blue}{(3+0+0)}$ counts 
$\color{red}{1} \times \color{blue}{3}$ possibilities.
$\color{red}{(1+1)}+\color{blue}{(2+1+0)}$ counts 
$\color{red}{1} \times \color{blue}{6}$ possibilities.
$\color{red}{(1+1)}+\color{blue}{(1+1+1)}$ counts 
$\color{red}{1} \times \color{blue}{1}$ possibility.
Totally $35$ possibilities.
