Let $n$ three-digit numbers satisfy the following properties:

(1) No number contains the digit 0.

(2) The sum of the digits of each number is 9.

(3) The units digits of any two numbers are different.

(4) The tens digits of any two numbers are different.

(5) The hundreds digits of any two numbers are different.

Find the largest possible value of $n$.
Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0. We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$'s (which means go to the next digit). Then for example 324 can be obtained from 111 by the string AAGAGAAA. There are in total

$$
\frac{8!}{6! \cdot 2!} = 28
$$

such words, so $S$ contains 28 numbers. Now, from the conditions (3), (4), (5), if $\overline{a b c}$ is in $T$ then each of the other numbers of the form $\overline{* c}$ cannot be in $T$, neither $\overline{* *}$ can be, nor $\overline{a * *}$. Since there are $a+b-2$ numbers of the first category, $a+c-2$ from the second and $b+c-2$ from the third one. In these three categories there are

$$
(a+b-2) + (b+c-2) + (c+a-2) = 2(a+b+c) - 6 = 2 \cdot 9 - 6 = 12
$$

distinct numbers that cannot be in $T$ if $\overline{a b c}$ is in $T$. So, if $T$ has $n$ numbers, then $12n$ are the forbidden ones that are in $S$, but each number from $S$ can be a forbidden number no more than three times, once for each of its digits, so

$$
n + \frac{12n}{3} \leq 28 \Longleftrightarrow n \leq \frac{28}{5}
$$

and since $n$ is an integer, we get $n \leq 5$. A possible example for $n=5$ is

$$
T = \{144, 252, 315, 423, 531\}
$$

Comment by PSC. It is classical to compute the cardinality of $S$ and this can be done in many ways. In general, the number of solutions of the equation

$$
x_{1} + x_{2} + \cdots + x_{k} = n
$$

in positive integers, where the order of $x_{i}$ matters, is well known that equals to $\binom{n-1}{k-1}$. In our case, we want to count the number of positive solutions to $a + b + c = 9$. By the above, this equals to $\binom{9-1}{3-1} = 28$. Using the general result above, we can also find that there are $a+b-2$ numbers of the form $\overline{* c}$.