Question

Let $n$ and $k$ be positive integers such that $n\ge \frac{k(k+1)}{2}$. Find the number of solutions $(x_1,x_2,...,x_k),x_1\ge 1,x_2\ge 2,x_k\ge k$ all integers satisfying $x_1+x_2+...+x_k=n$.Assume k=5 and n=15 .

Answer 1

We have $x_1+x_2+...+x_k=n...\left(1\right)$
Now let $y_1=x_1-1,y_2=x_2-2,...,y_k=x_k-k$
$\therefore y_1\ge 0,y_2\ge 0,....,y_k\ge 0$
Substituting the values $x_1,x_2,...,x_k$ in terms of $y_1,y_2,...,y_k$ in $\left(1\right)$, we have
$y_1+1+y_2+2+...+y_k+k=n$
$\Rightarrow y_1+y_2+...+y_k=n-(1+2+3+...+k)$
$\therefore y_1+y_2+...+y_k=n-\frac{k(k+1)}{2}=A\left(say\right)...\left(2\right)$
The number of non-negative integral solutions of the equation $\left(2\right)$ is
${}^{k+A-1}{C}_A=\frac{(k+A-1)!}{A!(k-1)!}$ where $A=n-\frac{k(k-1)}{2}$

Substituting values of $n$ and $k$, we get the required number as $126$