15. To prevent Xiaoqiang from being addicted to mobile games, Dad set a password on his phone. The phone password is 4 digits long, with each digit being a number between $0 \sim 9$. If the sum of the 4 digits used in the password is 20, Xiaoqiang would need to try at most $\qquad$ times to unlock the phone.
【Analysis】Consider the case of 20 being split into 4 numbers, where:
4 numbers are the same, there is 1 type: 5555
3 numbers are the same, there are 2 types: $6662, 4448$
2 types of numbers each appear 2 times, there are 4 types: $9911, 8822, 7733, 6644$
2 numbers are the same, the remaining are different, there are:
If the same number is 9, then the sum of the remaining 2 numbers is 2, there is only 20 this 1 type
If the same number is 8, then the sum of the remaining 2 numbers is 4, there are $40, 31$ these 2 types
If the same number is 7, then the sum of the remaining 2 numbers is 6, there are $60, 51, 42$ these 3 types
If the same number is 6, then the sum of the remaining 2 numbers is 8, there are $80, 71, 53$ these 3 types
If the same number is 5, then the sum of the remaining 2 numbers is 10, there are $91, 82, 73, 64$ these 4 types
If the same number is 4, then the sum of the remaining 2 numbers is 12, there are $93, 75$ these 2 types
If the same number is 3, then the sum of the remaining 2 numbers is 14, there are $95, 86$ these 2 types
If the same number is 2, then the sum of the remaining 2 numbers is 16, there is only 97 this 1 type
If the same number is 1, then the sum of the remaining 2 numbers is 18, there are 0 types, a total of 18 types
4 numbers are all different, there are 16 types:
$$
\begin{aligned}
20 & =9+8+3+0=9+8+2+1=9+7+4+0=9+7+3+1=9+6+5+0 \\
& =9+6+4+1=9+6+3+2=9+5+4+2=8+7+5+0=8+7+4+1 \\
& =8+7+3+2=8+6+5+1=8+6+4+2=8+5+4+3=7+6+5+2 \\
& =7+6+4+3
\end{aligned}
$$
4 numbers are the same, there is only 1 arrangement
3 numbers are the same, there are $C_{4}^{1}=4$ arrangements
2 types of numbers each appear 2 times, there are $C_{4}^{2}=6$ arrangements
2 numbers are the same, the remaining are different, there are $P_{4}^{2}=12$ arrangements
4 numbers are all different, there are $P_{4}^{4}=24$ arrangements
In summary, the password has $1 \times 1+2 \times 4+4 \times 6+18 \times 12+16 \times 24=633$ possibilities, meaning it can be unlocked with at most 633 attempts.