Find the value of $$\mathop{\sum\sum\sum\sum}_{0\leq i \leq j \leq k \leq l\leq n} 1$$
I am not sure but perhaps the answer is ${n+5}\choose 5$.
We know that $\displaystyle\mathop{\sum\sum\sum\sum}_{0\leq i < j < k < l\leq n} 1 $ is ${n+1}\choose 4$. Can we do this problem from this fact using the Principle of Inclusion-Exclusion?
Solutions using bijection would be quite elegant as well
This is the number of quadruples $(i,j,k,l)$ with $0\le i\le j\le k\le l$. Let $(i',j',k',l')=(i+1,j+2,k+3,l+4)$. Then the $(i',j',k',l')$ are the quadruples with $0<i'<j'<k'<l'\le n+4$. How many of these are there?