A permutation $\sigma$ of the numbers $1,2,\ldots , 10$ is called $\textit{bad}$ if there exist integers $i, j, k$ which satisfy
\[1 \leq i < j < k \leq 10 \quad \text{ and }\quad \sigma(j) < \sigma(k) < \sigma(i)\]
and $\textit{good}$ otherwise. Find the number of $\textit{good}$ permutations.
1. **Understanding the Problem:**
   We need to find the number of permutations of the numbers \(1, 2, \ldots, 10\) that do not contain a subsequence of the form \( \sigma(j) < \sigma(k) < \sigma(i) \) for \(1 \leq i < j < k \leq 10\). Such permutations are known as 312-avoiding permutations.

2. **Characterizing 312-Avoiding Permutations:**
   A permutation is 312-avoiding if it does not contain any subsequence of three elements where the first element is the largest, the second element is the smallest, and the third element is in between. These permutations are also known as stack-sortable permutations.

3. **Catalan Numbers:**
   The number of 312-avoiding permutations of \(n\) elements is given by the \(n\)-th Catalan number. The \(n\)-th Catalan number \(C_n\) is given by:
   \[
   C_n = \frac{1}{n+1} \binom{2n}{n}
   \]
   This formula arises from the recursive nature of 312-avoiding permutations.

4. **Recursive Definition:**
   Let \(g(n)\) denote the number of 312-avoiding permutations of \(1, 2, \ldots, n\). The recursive formula for \(g(n)\) is:
   \[
   g(n) = \sum_{j=0}^{n-1} g(j) g(n-1-j)
   \]
   This recurrence relation is characteristic of Catalan numbers.

5. **Calculating \(g(10)\):**
   Using the formula for Catalan numbers, we calculate \(C_{10}\):
   \[
   C_{10} = \frac{1}{10+1} \binom{20}{10} = \frac{1}{11} \binom{20}{10}
   \]
   First, compute \(\binom{20}{10}\):
   \[
   \binom{20}{10} = \frac{20!}{10! \cdot 10!}
   \]
   Using factorial values:
   \[
   20! = 2,432,902,008,176,640,000
   \]
   \[
   10! = 3,628,800
   \]
   Therefore:
   \[
   \binom{20}{10} = \frac{2,432,902,008,176,640,000}{(3,628,800)^2} = 184,756
   \]
   Now, compute \(C_{10}\):
   \[
   C_{10} = \frac{1}{11} \times 184,756 = 16,796
   \]

The final answer is \(\boxed{16,796}\).