Given a permutation $\sigma = (a_1,a_2,a_3,...a_n)$ of $(1,2,3,...n)$ , an ordered pair $(a_j,a_k)$ is called an inversion of $\sigma$ if $a \leq j < k \leq n$ and $a_j > a_k$. Let $m(\sigma)$ denote the no. of inversions of the permutation $\sigma$. Find the average of $m(\sigma)$ as $\sigma$ varies over all permutations.
To find the average number of inversions \( m(\sigma) \) over all permutations \(\sigma\) of \((1, 2, 3, \ldots, n)\), we can proceed as follows:

1. **Definition of Inversion**:
   An inversion in a permutation \(\sigma = (a_1, a_2, \ldots, a_n)\) is a pair \((a_j, a_k)\) such that \(1 \leq j < k \leq n\) and \(a_j > a_k\).

2. **Total Number of Permutations**:
   The total number of permutations of \(n\) elements is \(n!\).

3. **Counting Inversions**:
   For each pair \((j, k)\) with \(1 \leq j < k \leq n\), we need to determine the probability that \(a_j > a_k\). Since \(a_j\) and \(a_k\) are equally likely to be any of the \(n\) elements, the probability that \(a_j > a_k\) is \(\frac{1}{2}\).

4. **Number of Pairs**:
   The number of pairs \((j, k)\) with \(1 \leq j < k \leq n\) is given by the binomial coefficient:
   \[
   \binom{n}{2} = \frac{n(n-1)}{2}
   \]

5. **Expected Number of Inversions**:
   Since each pair \((j, k)\) has a \(\frac{1}{2}\) chance of being an inversion, the expected number of inversions in a permutation is:
   \[
   \mathbb{E}[m(\sigma)] = \binom{n}{2} \cdot \frac{1}{2} = \frac{n(n-1)}{2} \cdot \frac{1}{2} = \frac{n(n-1)}{4}
   \]

6. **Conclusion**:
   Therefore, the average number of inversions \(m(\sigma)\) over all permutations \(\sigma\) of \((1, 2, 3, \ldots, n)\) is:
   \[
   \boxed{\frac{n(n-1)}{4}}
   \]