12. Two schools compete in a chess tournament, each sending 75 students: 75 matches are organized to have all students play one against one (one member of the first school against one of the second) under the supervision of an external referee. Each school numbers its students from 1 to 75, and the referee imposes the rule that two boys cannot face each other if the difference between their numbers is a multiple of 3. If $n$ is the number of possible pairings that satisfy this rule, how many zeros does $n$ end with?
(A) 6
(B) 12
(C) 15
(D) 18
(E) 21

## Numerical answer problems -5 points
12. The answer is (D). Divide the 75 students of each of the two schools into 3 groups, depending on the remainder of their number when divided by 3. In particular, in the first school, the students are divided into groups $A, B, C$ depending on whether their number is of the form $3k, 3k+1$ or $3k+2$ for some integer $k$. Similarly, in the second school, the students are divided into $A', B', C'$.

Each of the groups $A, B, C, A', B', C'$ comprises 25 students, and the group $A$ cannot challenge the group $A'$, just as $B$ cannot challenge $B'$ and $C$ cannot challenge $C'$.

Suppose that $m$ people from group $A$ challenge someone in $B'$ and the remaining $25-m$ challenge $C'$. Then $m$ people from $B$ challenge the remaining $m$ in $C'$ and $25-m$ challenge students in $A'$. Consequently, there are $m$ people in $C$ who challenge the group $A'$ and the remaining $25-m$ challenge the group $B'$. For a fixed $m$, therefore, we choose $m$ people from $A, B, C$ in $\left(\begin{array}{c}25 \\ m\end{array}\right)^{3}$ ways. Additionally, we choose how the people in $A'$ challenge the $m$ chosen in $C$ and the $25-m$ in $B$ in 25 ! ways, and similarly for $B', C'$. As $m$ varies, therefore, the number of possible pairings is

$$
n=(25 !)^{3} \sum_{m=0}^{25}\left(\begin{array}{l}
25 \\
m
\end{array}\right)^{3}
$$

Notice that $\left(\begin{array}{c}25 \\ m\end{array}\right)$ is divisible by 5 exactly if $m \neq 0$ and $m \neq 25$, so

$$
\left(\begin{array}{l}
25 \\
m
\end{array}\right)^{3}=2+5 h
$$

is not a multiple of 5. The number $n$ therefore has as many factors of 5 as there are in $(25 \text { ! })^{3}$, which is $6 \cdot 3=18$. Notice that in 25 ! there are at least 6 factors of 2, for example because it includes $16 \cdot 8$, and therefore $n$ ends exactly with 18 zeros.