How many ten letter words are there with no repeated letters that contain neither the word ERGO nor the word LATER? How many ten letter words are there with no repeated letters that contain neither the word ERGO nor the word LATER?
I am thinking that there are 26^10 words with ten letters and 26P10 10 letter words with repeated lettes. 
So I would begin by taking 26^10 - 26P10, but not sure how to subtract the words containing ERGO or LATER.  Also not sure what to add to the end to account for the words that repeat and contain ERGO, the words that repeat and contain LATER, the words that contain ERGO and LATER, and the words that contain repeated letters and contain ERGO and LATER.
 A: Given an alphabet of $n$ letters one can form $$n\cdot(n-1)\cdots(n-r+1)={n!\over (n-r)!}$$ words of length $r$ using no letter twice.
It follows that there are ${26!\over16!}$ such words of length $10$ from the English alphabet.
There are ${22!\over16!}$ such words of length $6$ not using the letters occurring in ERGO. Each of these words has $7$ spaces (including the ends) where you can insert ERGO.
There are ${21!\over16!}$ such words of length $5$ not using the letters occurring in LATER. Each of these words has $6$ spaces (including the ends) where you can insert LATER.
There are ${19!\over16!}$ such words of length $3$ not using the letters occurring in LATERGO. Each of these words has $4$ spaces (including the ends) where you can insert LATERGO. 
Apart from the words containing LATERGO there are no $10$-letter words using no letter twice and containing ERGO as well as LATER.
Using the inclusion-exclusion principle we therefore obtain the following number of words containing no letter twice and not containing ERGO or LATER:
$${26!\over16!}-{22!\over16!}\cdot 7-{21!\over16!}\cdot 6+{19!\over16!}\cdot 4=19\,274\,833\,290\,456\ .$$
A: all words with no repetetion: $P\left( 26,10 \right)$
words including ERGO: forms are like  (ERGO_ _ _ _ _ _ ) or (_ _ ERGO _ _ _ _ )
$P\left( 22,6 \right)\cdot 7$ words.
words including LATER: similarly
$P\left( 21,5 \right)\cdot 6$ words.
including both: ERGOLATER_
$P\left( 3,3 \right)$ words.
now use inclusion-exclusion principle
$P\left( 26,10 \right)-\left[ P\left( 22,6 \right)\cdot 7+P\left( 21,5 \right)\cdot 6\text{ }\!\!~\!\!\text{ } \right]+P\left( 3,3 \right)$
