Question 188, Let $a_{1}, a_{2}, \ldots, a_{21}$ be a permutation of $1, 2, \ldots, 21$, satisfying $\left|a_{20}-a_{21}\right| \geq$ $\left|a_{19}-a_{21}\right| \geq \cdots \geq\left|a_{1}-a_{21}\right|$, the number of such permutations is $\qquad$ —.
Question 188, Answer: The answer is 3070.
For a given $k \in\{1,2, \ldots, 21\}$, we seek the number of permutations that satisfy the condition $a_{21}=k$.
(1) When $k \in\{1,2, \ldots, 11\}$, it must be that $\left\{a_{1}, a_{2}\right\}=\{k+1, k-1\},\left\{a_{3}, a_{4}\right\}=\{k+2, k-2\}, \cdots,\left\{a_{2 k-3}, a_{2 k-2}\right\}=\{2 k-1,1\}, a_{2 k-1}=2 k, a_{2 k}=2 k+1, \cdots, a_{20}=21$. Therefore, the number of such permutations is $2^{k-1}$.
(2) When $k \in\{12,13, \ldots, 21\}$, it must be that $\left\{a_{1}, a_{2}\right\}=\{k+1, k-1\},\left\{a_{3}, a_{4}\right\}=\{k+2, k-2\}, \cdots,\left\{a_{2(21-k)-1}, a_{2(21-k)}\right\}=\{21,2 k-21\}, a_{2(21-k)+1}=2 k-22, a_{2(21-k)+2}=2 k-23, \cdots, a_{20}=1$. Therefore, the number of such permutations is $2^{21-k}$.
Thus, the total number of permutations that satisfy the condition is: $\sum_{k=1}^{11} 2^{k-1}+\sum_{k=12}^{21} 2^{21-k}=3070$.