Arrangements of letters using inclusion and exclusion principles

Question

Using inclusion exclusion principles,

How many arrangements of the $26$ different letters are there that contain either the sequence "the" or the sequence "aid"

Here are my workings, are they correct? or where did I go wrong?

let,

$N(A)=$ number of arrangements with the sequence "the"
$N(B)=$ number of arrangements with the sequence "aid"
$N(A\bigcap B)=$ number of arrangemtns with the sequence "the" and "aid"

Calculating $N(A)$:
$26$ letters in the alphabet, consider the sequence "the" as one super letter, leaving $23$ other letters. Then the total number or arrangements would be $24!$ (which is the super letter + remaining letters)

Calculating $N(B)$:
$26$ letters in the alphabet, consider the sequence "aid" as one super letter, leaving $23$ other letters. Then the total number or arrangements would be $24!$ (which is the super letter + remaining letters)

Calculating $N(A\bigcap B)$:
$26$ letters in the alphabet, consider the sequence "the" and "aid" as two super letters, leaving $20$ other letters. Then the total number or arrangements would be $22!$ (which is the super letters + remaining letters)

Then $N(A) + N(B) - N(A\bigcap B) = 24! + 24! - 22!$

Answer 1

As already stated in the comments, your solution is correct. Your problem statement is a bit unclear, though, as “either/or” can be taken to mean “one or the other but not both”, which is apparently not what you meant.