Let $n$ and $k$ be positive integers such that $n\ge\frac{k(k+1)}{2}$. The number of solutions $(x_1,x_2,\dots,x_{k})$, with $x_1\ge1$, $x_2\ge2$,..., $x_{k}\ge k$ for all integers satisfying $x_1+x_2+\dots+x_{k}=n$ is?
I substituted the last equation in the first inequality. $$x_1+x_2+\dots+x_{k}\ge\frac{k(k+1)}{2}.$$ Took $x_1$ as $1+ t_1$, $x_2$ as $2+t_2$...where $t_i\ge 0$. On simplifying by using sum of k numbers, I end with with $t_1+t_2+\dots+t_{k} \ge0$. Since $x_1,x_2$... are in increasing order, and sum of all $t$ values is $0$, I conclude that this is only possible when $t=0$. Therefore only one solution is possible when $LHS = RHS$. The inequality is not valid.
But the answer is $\frac{1}{2}(2n-k^2+k-2)$. What am I missing here?