Let $n$, $(n \geq 3)$ be a positive integer and the set $A$={$1,2,...,n$}. All the elements of $A$ are randomly arranged in a sequence $(a_1,a_2,...,a_n)$. The pair $(a_i,a_j)$ forms an $inversion$ if $1 \leq i \leq j \leq n$ and $a_i > a_j$. In how many different ways all the elements of the set $A$ can be arranged in a sequence that contains exactly $3$ inversions?
To solve the problem, we need to determine the number of permutations of the set \( A = \{1, 2, \ldots, n\} \) that contain exactly 3 inversions. We will use the concept of generating functions and the recurrence relations for the number of permutations with a given number of inversions.

1. **Generating Function for Inversions:**
   The generating function for the number of inversions in permutations of \( S_n \) is given by:
   \[
   \sum_{\sigma \in S_n} x^{\text{inv}(\sigma)} = (1 + x)(1 + x + x^2) \cdots (1 + x + x^2 + \ldots + x^{n-1}).
   \]
   This generating function encodes the number of permutations with a given number of inversions.

2. **Recurrence Relations:**
   Let \( a_k(n) \) be the number of permutations in \( S_n \) with exactly \( k \) inversions. From the generating function, we can derive the following recurrence relations:
   \[
   a_1(n) = n - 1,
   \]
   \[
   a_2(n) = a_2(n-1) + a_1(n-1),
   \]
   \[
   a_3(n) = a_3(n-1) + a_2(n-1) + a_1(n-1).
   \]

3. **Base Cases:**
   We need to establish the base cases for the recurrence relations:
   \[
   a_1(2) = 1,
   \]
   \[
   a_2(3) = 1,
   \]
   \[
   a_3(4) = 1.
   \]

4. **Induction Hypothesis:**
   By induction, we can generalize the formulas for \( a_2(n) \) and \( a_3(n) \):
   \[
   a_2(n) = \frac{1}{2}n(n-1) - 1,
   \]
   \[
   a_3(n) = \frac{1}{6}n(n^2 - 7).
   \]

5. **Verification for \( a_3(n) \):**
   We need to verify the formula for \( a_3(n) \) by induction:
   - For \( n = 3 \):
     \[
     a_3(3) = \frac{1}{6} \cdot 3 \cdot (3^2 - 7) = \frac{1}{6} \cdot 3 \cdot 2 = 1.
     \]
   - For \( n = 4 \):
     \[
     a_3(4) = \frac{1}{6} \cdot 4 \cdot (4^2 - 7) = \frac{1}{6} \cdot 4 \cdot 9 = 6.
     \]

6. **Conclusion:**
   The number of permutations of the set \( A = \{1, 2, \ldots, n\} \) that contain exactly 3 inversions is given by:
   \[
   a_3(n) = \frac{1}{6}n(n^2 - 7).
   \]

The final answer is \(\boxed{\frac{1}{6}n(n^2 - 7)}\).