1. The problem asks to find the number of interesting ordered quadruples $(p, q, r, s)$ of integers such that $1 \leq p < q < r < s \leq 10$. 2. Since $p, q, r, s$ are strictly increasing integers between 1 and 10 inclusive, this is a problem of counting combinations without repetition. 3. The number of ways to choose 4 distinct integers from the set $\{1, 2, \dots, 10\}$ in strictly increasing order is the number of 4-combinations of 10 elements. 4. The formula for combinations is $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ For $n=10$ and $k=4$, we calculate: $$\binom{10}{4} = \frac{10!}{4! \times 6!}$$ 5. Compute the factorial values: $$10! = 3628800$$ $$4! = 24$$ $$6! = 720$$ 6. Substitute and simplify: $$\binom{10}{4} = \frac{3628800}{24 \times 720} = \frac{3628800}{17280} = 210$$ 7. Therefore, there are $210$ interesting ordered quadruples $(p, q, r, s)$ with $1 \leq p < q < r < s \leq 10$. **Final answer:** $$\boxed{210}$$