I have an array of numbers lets call it $p$ , where $p[r]={k+r-1\choose k-1}$


I want to find the sum of all the elements of $p$ taken $n$ at a time .


$0\le r\le k$


For instance, for $k=3$ ,$n=2$ , i need


$$p[0]^2+p[1]^2+p[2]^2+p[3]^2+p[0]\cdot p[1]+p[0]\cdot p[2]+p[0]\cdot p[3]+p[1]\cdot p[2]+p[1]\cdot p[3]+\\+p[2]\cdot p[3]=\sum\_{i\le j}p[i]\cdot p[j]=\sum\_{i\le j}{k+i-1\choose k-1}{k+j-1\choose k-1}$$ which is $273$


Is there any way , i can solve this for any value of $k$ or $n$ without solving big equations like above


In other words , how to solve this summation?


$$\sum\_{0\le x\_1\le x\_2...\le x\_n \le n}^{}\binom{k+x\_1-1}{x\_1}\binom{k+x\_2-1}{x\_2}...\binom{k+x\_n-1}{x\_n}$$



Sum Of n numbers taken $k$ at a time, where numbers are of form $r\choose k$
I'm not sure why this received a downvote. You are asking for the
coefficient of $x^n$ in $F(x)=1/\prod\_{r=0}^k (1-{k+r-1\choose
k-1}x)$. Since the denominator of $F(x)$ has distinct explicit zeros,
one can explicitly find the partial fraction expansion of $F(x)$ and
then expand each term. For instance, when $k=3$ we get
 $$ F(x) = \frac{1}{(1-x)(1-3x)(1-6x)(1-10x)} $$
 $$ \ =
 \frac{250}{63(1-10x)}-\frac{18}{5(1-6x)}+\frac{19}{14(1-3x)}
 -\frac{1}{90(1-x)}. $$
The coefficient of $x^n$ is therefore
$\frac{250}{63}10^n-\frac{18}{5}6^n +\frac{19}{14}3^n-\frac{1}{90}$.


