Find all n-tuples $(x_1,x_2, \cdots , x_n)$ of nonnegative integers $x_1 +x_2+ \cdots + x_n = k$ satisfying $x_1 < x_2 < \cdots < x_n$

Question

The number of $n$-tuples for $x_1 +x_2+ \cdots + x_n = k$ can be easily counted by the formula $ k+n-1 \choose n-1$. But what would happen if there is a restriction given by $x_1 < x_2 < \cdots < x_n$? I am lost with the calculation and wondering if there is a closed form solution.

Answer 1

Let $n \ge 2, n \in \mathbb N$. $A_n$ denotes the number of positive integer solutions to the equation $$x_1+2x_2+...+nx_n=n^2.$$

Note that $A_n$ is also equal to the number of tuples $(y_1,y_2,\ldots,y_n)\in\mathbb{Z}^n$ such that $0<y_1<y_2< \ldots y_n$ and $y_1+\ldots +y_n=n^2$.

[Consider the bijection $y_1=x_n,y_2=x_n+x_{n-1}$,$\ldots$,$y_n=x_n+\ldots +x_1$.]

Then $$\frac{n^n(n-1)^{n-1}}{2^{n-1}\left(n!\right)^2}<A_n<\frac{n^{2n-1}}{\left(n!\right)^2}$$

The number of such tuples is less than $\dfrac{1}{n!}$ times the number of positive integer solutions to $z_1+\ldots +z_n=n^2$, which is ${n^2-1 \choose n-1}$, which is at most $\dfrac{n^{2(n-1)}}{(n-1)!}=\dfrac{n^{2n-1}}{n!}$—this gives the upper bound.