15. Let the ordered array $A=\left(a_{1}, a_{2}, \cdots, a_{10}\right)$ satisfy the following 4 conditions:
(1) $a_{1}, a_{2}, \cdots, a_{10}$ is a permutation of $1,2, \cdots, 10$;
(2) $a_{1}a_{3}, a_{4}>a_{5}, a_{6}>a_{7}, a_{8}>a_{9}$;
(4) There does not exist $1 \leqslant i<j<k \leqslant 10$, such that $a_{i}<a_{k}<a_{j}$. Find the number of such ordered arrays $A$.
First, consider the following problem: For 10 points given on a circle, how many different ways (referred to as the number of pairings) are there to pair these 10 points using 5 non-intersecting line segments?

Let the number of pairings for $2n$ points on a circle, paired using $n$ non-intersecting line segments, be denoted as $f_{n}$. It is easy to see that $f_{1}=1$.

When $n \geqslant 2$, let the $2n$ points be $A_{1}, A_{2}, \cdots, A_{2n}$. Suppose $A_{1}$ is connected to $A_{k}$. According to the problem, $A_{2}, \cdots, A_{k-1}$ and $A_{k+1}, \cdots, A_{2n}$ are both even numbers of points, so $k=2m$.

- When $m=1$, the number of pairings is $f_{n-1}$.
- When $m=2,3, \cdots, n-1$, the number of pairings is $f_{m-1} f_{n-m}$.
- When $m=n$, the number of pairings is $f_{n-1}$.

Thus, $f_{n}=f_{n-1}+f_{1} f_{n-2}+f_{2} f_{n-3}+\cdots+f_{n-1}$.

Calculating, we get $f_{1}=1, f_{2}=f_{1}+f_{1}=2, f_{3}=f_{2}+f_{1} f_{1}+f_{2}=5$,
$$
f_{4}=f_{3}+f_{1} f_{2}+f_{2} f_{1}+f_{3}=14, f_{5}=f_{4}+f_{1} f_{3}+f_{2} f_{2}+f_{3} f_{1}+f_{4}=42.
$$

Therefore, the number of pairings for 10 points on a circle, paired using 5 non-intersecting line segments, is 42. Next, we establish a one-to-one correspondence between the array $\left(a_{1}, a_{2}, \cdots, a_{10}\right)$ that meets the conditions and the pairing method for the 10 points.

Let $A_{1}, A_{2}, \cdots, A_{10}$ be 10 points arranged counterclockwise on a circle, and label them with the numbers $1,2, \cdots, 10$. If the array $\left(a_{1}, a_{2}, \cdots, a_{10}\right)$ satisfies the conditions, connect the points labeled with the numbers $a_{2i-1}$ and $a_{2i}$ into a line segment $(i=1,2,3,4,5)$. By the conditions met by the array, these line segments do not intersect, thus forming a pairing method.

In fact, if there are two intersecting line segments, let the endpoints of these segments correspond to the numbers
$a_{2i-1}, a_{2i}$ and $a_{2j-1}, a_{2j}$, as shown in Figure 1. Then the points corresponding to the numbers $a_{2i-1}, a_{2j-1}$,
$a_{2i}, a_{2j}$ are arranged counterclockwise on the circle, i.e., $a_{2i-1} < a_{2j-1} < a_{2i} < a_{2j}$. If $a_{2i-1} < a_{2j-1} < a_{2i} < a_{2j}$, then $2j-1 < 2i < 2j$, which contradicts the condition that $a_{2i-1} < a_{2i}$ and $a_{2j-1} < a_{2j}$. Therefore, the line segments do not intersect.

For the array to satisfy the conditions, it must also meet:
- $a_{2i-1} < a_{2i}$, for $i=1,2,3,4,5$, i.e., condition (1).
- $a_{2i} < a_{2i+1}$, for $i=1,2,3,4$, i.e., condition (2).
- $a_{2i-1} < a_{2i+1}$, for $i=1,2,3,4$, i.e., condition (3).
- For any $i < j < k$, $a_{i} < a_{k} < a_{j}$ does not hold, i.e., condition (4).

For condition (4), if $i < j < k$ and $a_{i} < a_{j}$, then $a_{i} < a_{k} < a_{j}$ does not hold. If $a_{i} < a_{j}$, then $a_{i} < a_{k} < a_{j}$ does not hold. Therefore, for any $i < j < k$, $a_{i} < a_{k} < a_{j}$ does not hold, i.e., the array satisfies condition (4).

In summary, the number of ordered arrays $A$ that meet the conditions is 42.