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# Cycles, Symmetry, and Counting | ISI MStat 2016 PSB | Problem 2

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

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.

## Problem

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 $1,2, \ldots, 10$.

## Solution

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 $$\frac{\sum_{i, j = 1; i \neq j}^{n} {2ij}}{n-1}$$

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

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

#### Edit 1:

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

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