How many different arrangements of $A, B, C, D$, and $E$ which do not contain $AB$ and do not contain $BE$ ?
My try:
Number of $5$ letters permutations =$5!=120$
Number of arrangements that contain $AB = 4!=24$
Number of arrangements that contain $BE = 4!=24$
So Answer = $120-48=72$
But the answer in the book is $78$
What is wrong with my solution?
The problem is that some arrangements contain both $AB$ and $BE$, and you have subtracted them twice.
These arrangements can be thought of as arrangements of $3$ objects, $C$, $D$ and $ABE$. There are $3!$ of them, which is the $6$ that you are missing.
In general, see the inclusion-exclusion principle.