Can we use inclusion and exclusion principles on sums $(\sigma)$? 
want: $$ \sum_{i,j,k=0}^n \frac{1}{3^{i+j+k}}$$  with $ i \neq j \neq k$

Our condition is $ i \neq j \neq q$
We can split this into three inequalities:
$$ i \neq j \tag{a}$$
$$ j \neq k \tag{b}$$
$$ k \neq i \tag{c}$$
Now to the sum is over the set of $(i,j,k)$ triples satisfying the three properties stated above(*). By inclusion exclusion we can split this into three parts:
$$ a \cup b \cup c  = a + b + c - a \cap b - a \cap c -b \cap c + a \cap b \cap c$$
Hence we can write:
$$ \sum_{a \cup b \cup c} =  \sum_{ a + b + c - a \cap b - a \cap c -b \cap c + a \cap b \cap c}= \sum_a + \sum_b + \sum_c - \sum_{a \cap c} - \sum_{ b \cap c} - \sum_{ b \cap c} + \sum_{ a \cap b \cap c}$$
Hence for our our given problem(**):
$$ \sum_{i \neq j \neq q}^n \frac{1}{3^{i+j+k} } =  \sum_{i \neq j}\frac{1}{3^{i+j+k} }  + \sum_{ j \neq k} \frac{1}{3^{i+j+k} } + \sum_{ k \neq i}\frac{1}{3^{i+j+k} }  - \sum_{i=k \neq j} \frac{1}{3^{i+j+k} } - \sum_{j=k \neq i}\frac{1}{3^{i+j+k} }  - \sum_{i=j \neq k}\frac{1}{3^{i+j+k} }  + \sum_{i=j=k}\frac{1}{3^{i+j+k} }  $$
Now since:
$$ 0 \leq (i,j,k) \leq (n,n,n)$$
The first three terms in the sum are equivalent, and so are the three terms after that:
$$ \sum_{i \neq j \neq q}^n \frac{1}{3^{i+j+k} }= 3 \sum_{i \neq j}\frac{1}{3^{i+j+k} }- 3 \sum_{i=k \neq j} \frac{1}{3^{i+j+k} }  +\sum_{i=j=k}\frac{1}{3^{i+j+k} }$$
And then it's easy to simplify

How do I justify splitting the summation using set theory results? It came intuitively to me but I don't think I've seen this property applied yet.
If this was done before, does it have a name?

*: The set conditions define a subset on the set of i,j,k triples
**: it is implicit that I am summing over the condition $(0,0,0) \leq (i,j,k) \leq (n,n,n)$ I'd have written this but it'd be too much clutter.
***: Intersection of conditions i.e: $(a) \cap (b)$ is equivalent to saying to comb the statements "set of i not equal to j and set of k not equal to j" their inspection is the set " set of i=k not equal to j"
 A: A few comments on the notation:

*

*$i\ne j\ne k$ means simply that $i\ne j$ and $j\ne k$. It does not imply $i\ne k$. So the triplet $(0,1,0)$ satisfies the condition. I am assuming you want to sum over the condition $i,j,k$ distinct i.e. $i\ne j\ne k\ne i$.


*You can't denote inequalities and the set of triplets that satisfy those inequalities by the same label $(a),(b)$ or $(c)$. Let $A,B,C$ denote the sets of triples $(i,j,k)\in\{0,1,...,n\}^3$ that satisfy the inequalities $(a),(b),(c)$ as you have defined above respectively.

Now onto the errors in your approach:


*Since the inequalities are connected by the $AND$ logical operator, i.e. all of them need to be satisfied simultaneously, we are summing over the triplets in $A\cap B\cap C$ and not $A\cup B\cup C$. So you want to find$$\sum_{(i,j,k)\in A\cap B\cap C}3^{-i-j-k}$$


*$A\cap B$ is not the set of $(i,j,k)$ that satisfy $i=k\ne j$. You require $i\ne j$ and $j\ne k$ but not $i=k$. So both $(0,1,2),(0,1,0)$ both belong to $A\cap B$. Similarly, $A\cap B\cap C$ is not the set of triplets with $i=j=k$, rather the set of triplets with $i\ne j\ne k\ne i$.
Your final expression should be$$\sum_{i\ne j\text{ or }j\ne k\text{ or }k\ne i}3^{-i-j-k}=3\sum_{i\ne j}3^{-i-j-k}−3\sum_{i\ne j\ne k}3^{-i-j-k}+\color{red}{\sum_{i\ne j\ne k\ne i}3^{-i-j-k}}$$and we are interested in finding the red term. I am not sure if this is particularly easier to evaluate than the original expression.

The original expression can be easily evaluated by converting it into a nested summation:$$\sum_{i\ne j\ne k\ne i}3^{-i-j-k}=\sum_{i=0}^n\sum_{i\ne j=0}^n\sum_{i,j\ne k=0}^n3^{-i-j-k}\\=\sum_{i=0}^n3^{-i}\sum_{i\ne j=0}^n3^{-j}\sum_{i,j\ne k=0}^n3^{-k}$$Focus on the innermost summation. We are summing powers of $3$ except $3^{-i}$ and $3^{-j}$. Also since $i\ne j, 3^{-i},3^{-j}$ are distinct terms. Thus,$$\sum_{i,j\ne k=0}^n3^{-k}=\left(\frac1{3^0}+\frac1{3^1}+...+\frac1{3^n}\right)-\frac1{3^i}-\frac1{3^j}=\frac32(1-3^{-n-1})-\frac1{3^i}-\frac1{3^j}$$Moving on to the middle summation,$$\sum_{i\ne j=0}^n\frac1{3^j}\left(\frac32(1-3^{-n-1})-\frac1{3^i}-\frac1{3^j}\right)=\left(\frac32(1-3^{-n-1})-\frac1{3^i}\right)\sum_{i\ne j=0}^n\frac1{3^j}-\sum_{i\ne j=0}^n\frac1{9^j}\\=\left(\frac32(1-3^{-n-1})-\frac1{3^i}\right)^2-\left(\frac98(1-9^{-n-1})-\frac1{9^i}\right)$$Opening the square term and summing over $i=0\to n$,$$\left[\frac94(1-3^{-n-1})^2-\frac98(1-9^{-n-1})\right]\sum_{i=0}^n\frac1{3^i}-3(1-3^{-n-1})\sum_{i=0}^n\frac1{9^i}+\sum_{i=0}^n\frac2{27^i}\\=\left[\frac94(1-3^{-n-1})^2-\frac98(1-9^{-n-1})\right]\frac32(1-3^{-n-1})-\frac{27}8(1-3^{-n-1})(1-9^{-n-1})+\frac{27}{13}(1-{27}^{-n-1})\\=\frac{27}8(1-3^{-n-1})^3-\frac{81}{16}(1-9^{-n-1})(1-3^{-n-1})+\frac{27}{13}(1-{27}^{-n-1})$$You can choose to further simplify but I will rest it here. You can verify that the summation tends to $81/208$ as $n\to\infty$.
