Mongolian Mathematical Olympiad

Let $A_0,A_1,A_2$ be subsets of $\{2,3,...,49\}$ whose elements give remainder $0$, $1$, $2$ after dividing by $3$. Note that $A_0\cup A_1\cup A_2=\{2,3,...,49\}$. Define a mapping $f:\{A_0,A_1,A_2\}\to \{0,1,2\}$ as follows: if $x\in A_i$ then we put $x=i$. Therefore, to form a sequence with the given condition, it is sufficient to arrange $1$'s and $2$'s so the sum of them is not divisible by $3$ and in the remaining places put $0$'s. Since $16+17=33$, consider the sequence $2,2,1,2,1,...,2,1$ and we need to arrange $16$ $0$'s between them. Observe that $S_1$ is not divisible by $3$. Therefore, we don't put $0$ in the $1$st place. Number of ways putting $0$'s in the remaining $48$ places is $\frac{48!}{32!\cdot 16!}$ and number of ways distributing elements of $A_0$ in a chosen place is $16!$. Thus, by the product principle, number of ways distributing elements of $A_0$ is $\frac{48!}{32!\cdot 16!}\cdot 16!=\frac{48!}{32!}$. In the sequence $2,2,1,2,1,...,2,1$ number of ways distributing $1$'s and $2$'s is $16!\cdot 17!$. Hence we have $\frac{48!}{32!}\cdot 16!\cdot 17!$.