How many divisors of $11$ are in a $9$ digit number containing distinct digits from $\{1,2,...9\}$ (no repetitition)
My attempt : Let $e$, $o$ denote sum of digits at even and odd places.
$e+o=1+2...9=45$
$|e-o|=0,11,22,33...$
Also,
$10\leqslant e\leqslant 30$
$15\leqslant o\leqslant 35$ As $e$ and $o$ are natural numbers, we see that only possible case is :
$|e-o|=11$
Either $e=28,o=17$ OR $e=17,o=28$
Let $e=a_2+a_4+a_6+a_8$
$o=a_1+a_3+a_5+a_7+a_9$
The total number of divisors are $(\alpha+\beta) 5! 4!$ where those two variables are solutions for two conditions obtained earlier.
What is the best strategy to find their number of solutions manually? I tried to related it with multinomial but that is too long.
Because $e$ is so close to the limit in both cases, enumerating by hand is not too hard. The highest $e$ can be is $30$ and for the first case you need it to be $28$. Starting with $9876$, we need to subtract $2$, and the only choices are $9874$ and $9865$. In the other case, the minimum $e$ can be is $10$ and you need $17$. There are more possibilities, but still within hand work being systematic. I find $1259,1268,1349,1358,1367,1457,2348,2357,2456$ for nine.
You could do this with stars and bars, but the numbers are small enough that hand work is easier.