How many arrangements of the digits $0,1,....,9$ are there that do not end with an $8$ and do not begin with a $3$?

Question

Using inclusion exclusion principles,

How many arrangements of the digits $0,1,....,9$ are there that do not end with an $8$ and do not begin with a $3$?

Here are my workings, are they correct? or where did I go wrong? I am not sure if I used to the intersection and union of $A$ and $B$ correctly

let,

$n(A)=$ number of arrangements ending with $8=9!$
$n(B)=$ number of arrangements beginning with $3=9!$
$n(E)=$ total number of arrangements $=10!$
$n(A\bigcap B)= 8!$

We require: $n(E)-n(A\bigcup B)$
but $n(A\bigcup B)= n(A) + n(B) - n(A\bigcap B)$
Since $A$ and $B$ are not mutually exclusive, so
$n(E) - (n(A)+n(B)-(n(A\bigcap B)) = 10!-(9!+9!-8!)$

Answer 1

Assuming no. beginning with 0 is valid,

no. of arrangements=$n(\text{total possibilities})-n(\text{no.s with }3\text{ at start and end with }8)$

$n=10!-8!$

Answer 2

There are $10!$ permutations at all. Now $1\cdot8\cdot7\cdot6\cdot5\cdot4\cdot4\cdot3\cdot2\cdot1$ start with a three and end with an eight. Now subtract.