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
 A: 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*}
A: There are $2^{5-1} -2^{(2-1)-1}=15$ compositions of $5$ into at least $2$ positive integers while there are $6$ partitions of $5$ into at least $2$ positive integers.  With compositions, $1+1+3$ is distinct from $3+1+1$, while with partitions they are the same.
But your first attempt counts even more than compositions.  It actually counts cases where the first two elements are positive and the next three are non-negative, including $3+1+0+0+1$.  I doubt this is what you want to count.
A: 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.
