$\mathit Exact \ question :$
How many 7-digit numbers (without repetition) can be formed from digits
$$1,2,3,4,5,6,7,8,9 $$ such that each of them are divisible by $ 18 $ ?
$\mathit My \ approach :$
First I checked for the divisibility of $9$, the numbers would be...
$$ 1236789, 1245789 ,1345689, 2345679 $$
In this, at last position either of $2,4,6,8 $ can come, that leaves us to the total possibility of only 3 digits appearing at last position.
And the total 7-digit numbers would be
$$\displaystyle 3 \times (6! \times 4) = 8640. $$ Is it correct ? Actually, I came up with this question myself.