I'll give his one a try
We have 6 ways to choose a consonant for the first position and 5 remaining ways to choose a consonant for the last position
And the other letters can be arranged in 7! ways
However........we must divide this by [2! *2!] to count only the "identifiable" words since we have repeated "O's" and "S's"
So we have
[ 6 * 5 * 7! ] / [2! * 2!] = 37800 arrangements
[ Can some other mathematician (Melody??) check this ??? ]
