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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

Mathejunior
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1 Answers1

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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?

Angina Seng
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  • We can re-write the same thing as: $0\leq i'' < j'' < k'' < l'' \leq n+3$ given that $x''=x'-1~ \forall x\in{i,j,k,l}$. And thus, our answer becomes ${n+4}\choose 4$. Now, for $n=0$, it works, right? It indeed is $1$ as $i=j=k=l=0$ and that's the only choice we have. – Mathejunior Oct 28 '17 at 05:44
  • @Mathbg That's right! – Angina Seng Oct 28 '17 at 05:45
  • Anyway, as I was trying, it's hard to do using Inclusion-Exclusion. There would be a lot of cases to consider individually. – Mathejunior Oct 28 '17 at 05:48
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    @Mathbg: there is no need of inclusion-exclusion, this is just stars and bars (https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)) in disguise. – Jack D'Aurizio Oct 28 '17 at 15:59