We have the letters 'WINPRESENT'. I want to calculate the rearrangements of these letters that contain either the word 'WIN' or the word 'PRESENT' or both of them.
I have done the following:
The subword 'WIN' is contained in $8!$ rearrangements. The subword 'PRESENT' is contained in $4!$ rearrangements. We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.
So the total amount of rearrangements that contain either the word 'WIN' or the word 'PRESENT' or both of them is equal to $8!+4!-2$, or not?
This is a good approach, but not quite right: when considering the number of arrangements which contain WIN, you need to take account of the fact that there are two Es in the remaining letters. As it is, you've counted $\text{SPE}_1(\text{WIN})\text{NE}_2\text{TR}$ and $\text{SPE}_2(\text{WIN})\text{NE}_1\text{TR}$ as different arrangements.