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I thought of using the balls and flags trick - where you use the flags to separate identical balls into different groups. But I am not able to discard the solutions in which I get $zeroes$. Please provide with a simplified proof. Thank you! :)

Ishaan
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    Number of positive integer solutions to $x_1+x_2+...+x_k=n$ is $\binom{n-1}{k-1}$ by your 'balls and flags' method. There are so many posts here regarding this type of problem. Have a look at this. – StubbornAtom Nov 13 '16 at 14:26
  • Yeah, thanks! Sorry for re-posting the question. :( My bad! – Ishaan Nov 13 '16 at 14:30

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To emphasize what @StubbornAtom said : consider an array of $n$ ones. To generate a sum of $k$ numbers which add p to $n$, you just have to separate your array in $k$ packets, which requires $k-1$ "walls". There are \binom{n-1}{k-1}$ ways to choose the places of your walls.

Example : $x+y+z=7$ : $$(1+1+1+1+1+1+1) \longrightarrow (1+1|1+1+1|1+1) \longrightarrow 2+3+2$$ You have to change $2=k-1$ of the $6=n-1$ "$+$" into "$|$".

Nicolas FRANCOIS
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