Consider the following array:
\[ 3, 5\\3, 8, 5\\3, 11, 13, 5\\3, 14, 24, 18, 5\\3, 17, 38, 42, 23, 5\\ \ldots
\] Find the 5-th number on the $ n$-th row with $ n>5$.
1. To solve the problem, we need to find the 5th number on the \( n \)-th row for \( n > 5 \). The solution involves understanding the pattern in the given array and the relationship with Pascal's triangle.

2. The given array is:
   \[
   3, 53, 8, 53, 11, 13, 53, 14, 24, 18, 53, 17, 38, 42, 23, 5 \ldots
   \]

3. The solution suggests that the array can be derived by summing a modified Pascal's triangle. Let's break down the steps to understand this.

4. The modified Pascal's triangle is given as:
   \[
   \begin{array}{cccccc}
   2 & 4 \\
   2 & 6 & 4 \\
   2 & 8 & 10 & 4 \\
   2 & 10 & 18 & 14 & 4 \\
   2 & 12 & 28 & 32 & 18 & 4 \\
   \end{array}
   \]

5. This triangle is obtained by summing another triangle:
   \[
   \begin{array}{cccccc}
   1 & 2 \\
   1 & 3 & 2 \\
   1 & 4 & 5 & 2 \\
   1 & 5 & 9 & 7 & 2 \\
   1 & 6 & 14 & 16 & 9 & 4 \\
   \end{array}
   \]

6. The second triangle is obtained by summing Pascal's triangle:
   \[
   \begin{array}{cccccc}
   0 & 1 \\
   0 & 1 & 1 \\
   0 & 1 & 2 & 1 \\
   0 & 1 & 3 & 3 & 1 \\
   0 & 1 & 4 & 6 & 4 & 1 \\
   \end{array}
   \]

7. The desired formula for the 5th number on the \( n \)-th row is given by:
   \[
   \frac{n(n-1)(n-2)(n-3)}{8} + \frac{(n-1)(n-2)(n-3)}{3}
   \]

8. Let's simplify the formula step-by-step:
   \[
   \frac{n(n-1)(n-2)(n-3)}{8} + \frac{(n-1)(n-2)(n-3)}{3}
   \]

9. Combine the terms over a common denominator:
   \[
   \frac{3n(n-1)(n-2)(n-3) + 8(n-1)(n-2)(n-3)}{24}
   \]

10. Factor out the common term \((n-1)(n-2)(n-3)\):
    \[
    \frac{(n-1)(n-2)(n-3)(3n + 8)}{24}
    \]

11. Therefore, the 5th number on the \( n \)-th row is:
    \[
    \frac{(n-1)(n-2)(n-3)(3n + 8)}{24}
    \]

The final answer is \( \boxed{ \frac{(n-1)(n-2)(n-3)(3n + 8)}{24} } \).