Question

Number of $9$ digits numbers divisible by nine using the digits from $0$ to $9$ if each digit is used almost once is $K=8!$, then K has the value equal to____

Answer 1

$1+2+...3...9=\frac{9\times 10}{2}=95$

All the $9$ digit numbers using $1$ to $9$ are divisible by $9$

$9!$
The case when we use $0$ we must exclude $9$ from the number also $0$ can't be at the $9^{th}$
$8\times 8!$
$9!+8\times 8!=12\times 8!$
$K=17$