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This problem is a beautiful and elegant application of basic counting principles, symmetry and double counting principles in combinatorics. This is Problem 2 from ISI MStat 2016 PSB.

Determine the average value of

$$

i_{1} i_{2}+i_{2} i_{3}+\cdots+i_{9} i_{10}+i_{10} i_{1}

$$ taken over all permutations \(i_{1}, i_{2}, \ldots, i_{10}\) of \(1,2, \ldots, 10\).

- Permutation and Combination
- Symmetry Argument in Counting

The problem may seem mind boggling at first, when you will even try to do it for \(n = 4\), instead of \(n = 10\).

But, in mathematics, symmetry is really intriguing. Let's see how a symmetry argument holds here. It is just by starting to count. Let's see this problem in a geometrical manner.

\( i_{1}\to i_{2} \to i_{3} \to \cdots \to i_{9} \to i_{10} \to i_{1}\) is sort of a cycle right?

Now, the symmetry argument starts from this symmetric figure.

We will do the problem for general \(n\).

**Central Idea:** Let's fix a pair say [ 4 - 5 ], we will see in all the permutations, in how many times, [ 4 - 5 ] can occur.

We will see that there is nothing particular about [ 4 - 5 ], and this is the symmetry argument.Therefore, the number is symmetric along with all such pairs.

Observe, along every such cycle containing [ 4 - 5 ], there are three parameters:

- The position of the [ 4 - 5 ] in
**which edge**of the cycle? - The permutation of that [ 4 - 5 ], as
**[ 4 - 5 ] or [ 5 - 4 ]**. - The number of arrangements of the
**remaining \( n - 2\) numbers**.

The number corresponding to the above questions are the following:

- \(n\) options of position of edge since, there are \(n\) edges.
- 2! = 2 ways of arranging.
- \( (n-2)! \) ways of arranging the rest of the \( n -2\) numbers.

So, in total [ 4 - 5 ] edge will occur \( n \times 2! \times (n-2)! \) times.

By the symmetry argument, every edge [\( i - j\)], will occur \( n \times 2! \times (n-2)! \) times.

Thus, when we sum over all such permutations, we get the following

$$ \sum_{i, j = 1; i \neq j}^{n} {ij} \times \text{number of times [i - j] pair occur} $$

$$ = \sum_{i, j = 1; i \neq j}^{n} {ij} \times n \times 2! \times (n-2)! = n \times (n-2)! \times \sum_{i, j = 1; i \neq j}^{n} {2ij} $$

Now, there are \( n!\) permutations in total. So, to take the average, we divide by \( n!\) to get $$ \frac{\sum_{i, j = 1; i \neq j}^{n} {2ij}}{n-1} $$

$$ = \frac{(\sum_{i}^{n} i)^2 - \sum_{i}^{n} i^2 }{n-1} $$

$$ = \frac{\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}}{n-1} = \frac{n(n+1)(3n+2)}{12} $$

One of the readers, Vishal Routh has shared his solution using Conditional Expectation, I am sharing his solution in picture format.

This problem is a beautiful and elegant application of basic counting principles, symmetry and double counting principles in combinatorics. This is Problem 2 from ISI MStat 2016 PSB.

Determine the average value of

$$

i_{1} i_{2}+i_{2} i_{3}+\cdots+i_{9} i_{10}+i_{10} i_{1}

$$ taken over all permutations \(i_{1}, i_{2}, \ldots, i_{10}\) of \(1,2, \ldots, 10\).

- Permutation and Combination
- Symmetry Argument in Counting

The problem may seem mind boggling at first, when you will even try to do it for \(n = 4\), instead of \(n = 10\).

But, in mathematics, symmetry is really intriguing. Let's see how a symmetry argument holds here. It is just by starting to count. Let's see this problem in a geometrical manner.

\( i_{1}\to i_{2} \to i_{3} \to \cdots \to i_{9} \to i_{10} \to i_{1}\) is sort of a cycle right?

Now, the symmetry argument starts from this symmetric figure.

We will do the problem for general \(n\).

**Central Idea:** Let's fix a pair say [ 4 - 5 ], we will see in all the permutations, in how many times, [ 4 - 5 ] can occur.

We will see that there is nothing particular about [ 4 - 5 ], and this is the symmetry argument.Therefore, the number is symmetric along with all such pairs.

Observe, along every such cycle containing [ 4 - 5 ], there are three parameters:

- The position of the [ 4 - 5 ] in
**which edge**of the cycle? - The permutation of that [ 4 - 5 ], as
**[ 4 - 5 ] or [ 5 - 4 ]**. - The number of arrangements of the
**remaining \( n - 2\) numbers**.

The number corresponding to the above questions are the following:

- \(n\) options of position of edge since, there are \(n\) edges.
- 2! = 2 ways of arranging.
- \( (n-2)! \) ways of arranging the rest of the \( n -2\) numbers.

So, in total [ 4 - 5 ] edge will occur \( n \times 2! \times (n-2)! \) times.

By the symmetry argument, every edge [\( i - j\)], will occur \( n \times 2! \times (n-2)! \) times.

Thus, when we sum over all such permutations, we get the following

$$ \sum_{i, j = 1; i \neq j}^{n} {ij} \times \text{number of times [i - j] pair occur} $$

$$ = \sum_{i, j = 1; i \neq j}^{n} {ij} \times n \times 2! \times (n-2)! = n \times (n-2)! \times \sum_{i, j = 1; i \neq j}^{n} {2ij} $$

Now, there are \( n!\) permutations in total. So, to take the average, we divide by \( n!\) to get $$ \frac{\sum_{i, j = 1; i \neq j}^{n} {2ij}}{n-1} $$

$$ = \frac{(\sum_{i}^{n} i)^2 - \sum_{i}^{n} i^2 }{n-1} $$

$$ = \frac{\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}}{n-1} = \frac{n(n+1)(3n+2)}{12} $$

One of the readers, Vishal Routh has shared his solution using Conditional Expectation, I am sharing his solution in picture format.

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