[Combinatorics]
Problem: A permutation of a list of numbers is an ordered arrangement of the numbers in that list. For example, $3,2,4,1,6,5$ is a permutation of $1,2,3,4,5,6$. We can write this permutation as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, where $a_{1}=3, a_{2}=2, a_{3}=4, a_{4}=1, a_{5}=6$, and $a_{6}=5$.
Determine the average value of

$$
\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|
$$

over all permutations $a_{1}, a_{2}, a_{3}, a_{4}$ of $1,2,3,4$.

Solution: There are 4 ! $=4 \cdot 3 \cdot 2 \cdot 1=24$ permutations of $1,2,3,4$.

This is because there are 4 possible choices for $a_{1}$, and for each of these there are 3 possible choices for $a_{2}$, and for each of these there are 2 possible choices for $a_{3}$, and then 1 possible choice for $a_{4}$.

Consider the permutation $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4$. (We write this as $1,2,3,4$.)

Here, $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|=|1-2|+|3-4|=1+1=2$.

This value is the same as the value for each of $2,1,3,4$ and $1,2,4,3$ and $2,1,4,3$ and $3,4,1,2$ and 4,3,1,2 and 3,4,2,1 and 4,3,2,1.

Consider the permutation $1,3,2,4$.

Here, $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|=|1-3|+|2-4|=2+2=4$.

This value is the same as the value for each of $3,1,2,4$ and $1,3,4,2$ and $3,1,4,2$ and $2,4,1,3$ and 4,2,1,3 and 2,4,3,1 and 4,2,3,1.

Consider the permutation $1,4,2,3$.

Here, $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|=|1-4|+|2-3|=3+1=4$.

This value is the same as the value for each of 4,1,2,3 and 1,4,3,2 and 4,1,3,2 and 2,3,1,4 and $3,2,1,4$ and $2,3,4,1$ and $3,2,4,1$.

This accounts for all 24 permutations.

Therefore, the average value is $\frac{2 \cdot 8+4 \cdot 8+4 \cdot 8}{24}=\frac{80}{24}=\frac{10}{3}$.

Final answer: $\frac{10}{3}$