14. Let positive integers $a_{1}, a_{2}, \cdots, a_{10}$ be no greater than 21, and the sum of any two numbers is not equal to 21. Try to find the sum of the products $a_{1} a_{2} \cdots a_{10}$ of all arrays that satisfy the conditions.
Solve:
Group the positive integers $1 \sim 21$ as follows: $\{21\},\{1,20\},\{2,19\}, \cdots,\{10,11\}$,
If 21 is not taken, then $a_{1}, a_{2}, \cdots, a_{10}$ are taken from one of the last 10 sets, with a total of $2^{10} \cdot 10$ ! ways.
The corresponding product sum is $10!(1+20)(2+19) \cdots(10+11)=10!\cdot 21^{10}$;
If 21 is taken, for example, $a_{1}=21$, then the remaining 10 sets choose 9, with a total of $2^{10} \cdot 10$ ! ways.
The corresponding product sum is $10!(1+20)(2+19) \cdots(10+11)=10!\cdot 21^{10}$;
Similarly, the cases for $a_{2}=21, \cdots$ are the same.
In summary, the corresponding product sum is $10!\cdot 21^{10}+10 \cdot 10!\cdot 21^{10}=11!\cdot 21^{10}$.