安全代码

让我们以第一位数字为条件来确定代码的数量。

如果我们的第一个数字是 $9$，则有 $0$ 可能的代码（因为总和将超过非零数字的 $10$）

如果我们的第一位数字是$8$，那么剩下的数字之和只能是$2$。完成此操作的唯一方法是剩余数字为 $1$。

如果第一位数字是 $7$，则剩余的和可以是 $2$ 或 $3$。 $2$ 与 $3$ 的求和方式有多种（1+2 或 2+1），并且我们已经统计了与 $2$ 求和的方式，如果我们的第一个数字是 $7$，则总共有 $3$ 不同的组合

如果第一个数字是 $6$，我们只需要找到与 $4$ 相加的有多少种方法，并将其与我们之前看到的 $2$ 和 $3$ 的组合相加。 $3$ 有多种方法可以与 $4$ 求和。现在，我们应该注意到一个模式。

如果我们的第一个数字是$5$，就会有$1+2+3+4$不同的组合。本质上，我们只是对前一个 $8$ 三角数求和。 $$\begin{equation*} Total = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = \boxed{120} \end{equation*}$$

codes = []
for i in range(1, 10):
    for j in range(1, 10):
        for k in range(1, 10):
            codes.append([i, j, k])

valid_codes = [code for code in codes if sum(code) <= 10]
print(len(valid_codes))

Original Explanation

Let's condition the number of codes on the first digit.

If our first digit is $9$, there are $0$ possible codes (as sum will exceed $10$ with non-zero digits)

If our first digit is $8$, the remaining digits can only sum to $2$. The only way to accomplish this is if the remaining digits are $1$.

If the first digit is $7$, the remaining sum can be $2$ or $3$. There are $2$ ways to sum to $3$ (1+2 or 2+1) and we have already counted the ways to sum to $2$, resulting in a total of $3$ different combinations if our first digit is $7$

If the first digit is $6$ we only need to find how many ways to sum to $4$ and add that with the combinations for $2$ and $3$ we've seen previously. There are $3$ ways to sum to $4$. By now, we should notice a pattern.

If our first digit is $5$, there will be $1+2+3+4$ different combinations. Essentially, we are just summing the first $8$ triangular numbers.

codes = []
for i in range(1, 10):
    for j in range(1, 10):
        for k in range(1, 10):
            codes.append([i, j, k])

valid_codes = [code for code in codes if sum(code) <= 10]
print(len(valid_codes))