In how many ways can we arrange the numbers $1,2, \dots, 3n (n \geq 1)$, so that, at the positions that are multiple of $3$, there are only numbers that are multiple of $3$?
I thought that the answer is: $$\binom{3n}{n} \cdot n! \cdot (2n)!$$
Could you tell me if it is right?
Among the numbers from $1$ to $3n$, there are $n$ that are multiples of $3$, namely the numbers $3,6,9,\dots, 3n$. The count is easiest if we think of them as $3k$, where $k$ ranges from $1$ to $n$, so through $n$ values.
Similarly, there are $n$ positions that are multiples of $3$.
Since there are $n$ numbers divisible by $3$, and $n$ positions divisible by $3$, we have $n!$ ways to arrange the $n$ multiples of $3$ in allowed positions. For each of these ways, we can arrange the $2n$ remaining numbers in the remaining positions in $(2n)!$ ways.
Remark: We can change the problem and make it a little harder by asking that multiples of $3$ all land in positions that are not multiples of $3$. In that case, the set of positions occupied by multiples of $3$ can be chosen in $\binom{2n}{n}$ ways, and the rest goes as above.
