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:
$${26!\over16!}-{22!\over16!}\cdot 7-{21!\over16!}\cdot 6+{19!\over16!}\cdot 4=19\,274\,833\,290\,456\ .$$
Best Answer
Out of 5 vowel 1 vowel.
$\binom{5}{1}$
7 letters can out of 21 consonants can be choosen.
$21^7$
And 1 vowel can be on any 8 places in 8 ways.
Total = $\binom{5}{1} * 21^7 * 8$