Question: Let $n \geq 2$ be an even integer. A permutation $a_1; a_2; \ldots; a_n$ of the set $\{1,2, \ldots, n\}$ is called awesome if $a_2 = 2a_1$. For example, if $n = 6$, then the permutation $3; 6; 4; 1; 5; 2$ is awesome, whereas the permutation $3; 5; 4; 1; 6; 2$ is not awesome.
How many awesome permutations of the set $\{1,2, \ldots, n\}$ are there?
Answer: $\frac{n}{2} \cdot (n-2)!$
Attempt:
My understanding was since we need $a_2 = 2a_1$ then $a_1 = a_2/2$. So $a_1$ should be the form $n/2$. For $a_2$, I assumed since $a_1$ was already chosen to be $n$, then $a_2$ should be $n-1$. So the total permutations should be $(n/2) \cdot (n-1)!$
Best Answer
Not quite. $a_1$ should not necessarily be $\frac{n}{2}$; rather, it can be any number which is at most $\frac{n}{2}$. For example, $2,4,1,3,6,5$ would be awesome. So there's $\frac{n}{2}$ choices for $a_1$ in an awesome permutation, and once this is chosen, only one choice for $a_2$ (because it has to be $2a_1$). The rest of the $n-2$ numbers can be ordered arbitrarily in $(n-2)!$ ways, for a total of $\frac{n}{2} (n-2)!$ permutations.