Combinatorics-discrete mathematics

Question

Count the solutions in nonnegative integers $x_1,\ldots,x_k$ to $x_1 + \cdots + x_k \leq n$.

Can someone please help with the solution?

Answer 1

Add an extra variable $x_{k+1}$ and count the solutions in non-negative integers to the equation

$$x_1+x_2+\ldots+x_k+x_{k+1}=n\;;\tag{1}$$

this is a standard stars-and-bars problem, and the answer is $$\binom{n+(k+1)-1}{(k+1)-1}=\binom{n+k}k\;.$$ (The reasoning behind this is reasonably well explained in the linked article.) Every solution to $(1)$ gives you a solution to your inequality by simply throwing away $x_{k+1}$. On the other hand, if

$$x_1+x_2+\ldots+x_k\le n\;,$$

then you can set $x_{k+1}=n-(x_1+x_2+\ldots+x_k)$ to get a solution to $(1)$. Thus, there are exactly as many solutions to your inequality as there are solutions to $(1)$.

Answer 2

It seems that you can write

$0$ in $1$ way, $0 + 0 + \dots + 0$

$1$ in $k$ ways, $1 + 0 + 0 + \dots$, $0 + 1 + 0 + \dots$

$2$ in $k + {k \choose 2}$ ways, $2 + 0 + \dots$, $\dots$ and $1 + 1 + 0 + \dots$, $\dots$

$3$ in $k + {k \choose 2} + {k \choose 3}$ ways

Can you generalize and use this information to compute what you asked?

Answer 3

hint:if $ x_1 + ... + x_r+ = n$ then define this function $$f(x)=\sum_{r=0}^\infty b_r\frac{x^r}{r!} $$ and answer is coefficient of $\frac{x^r}{r!}$ in f(x)( f(x) is generate function) or answer is n!$n \choose k+1$ for this question at first add $x_{k+1}$ then we have $$x_1 + ... + x_k+ x_{k+1} = n$$ and according to above answer is : n!$n \choose k+1$