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

Solve the following question via inclusion and exclusion

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

My solution:

let,

$n(A)=$arrangements ending with $8 = 9!$
$n(B)=$arrangements ending with $3 = 9!$
$n(E)=$total 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:
$n(E)-(n(A)+n(B)-n(A\bigcap B))$

$= 10!-(9!+9!-8!)$

Is this the correct answer? Thanks everyone!

Answer 1

As already written in the comments, your approach and result are correct.