How many 6 letter words can be made with these conditions? The letters that can be used are A, I, L, S, T. 
The word must start and end with a consonant. Exactly two vowels must be used. The vowels can't be adjacent.
 A: There are $3$ ways to choose the first consonant, and $3$ ways to choose the last one.
Of the four remaining letters, exactly two must be vowels. There are $$\binom{4}{2}-3=6-3=3$$ ways to choose which spots the vowels can be in, since they cannot be adjacent. Then, there are $2$ choices for the first vowel, and $2$ choices for the second. 
Finally, there are $3$ ways to choose each of the remaining two consonants.

Multiplying all of these numbers together, we find the total number of $6$-letter words that fit all the constraints: $$3 \cdot 3 \cdot \left(\binom{4}{2}-3\right) \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 2^2 \cdot 3^5 = \boxed{972}$$
A: Since the words begin and end with a consonant we can see that there are three places to put the two vowels such that they are not adjacent. The three possible placings of the $6$ letter words are structured as $$3\cdot2\cdot3\cdot2\cdot3\cdot3=324,$$ $$3\cdot3\cdot2\cdot3\cdot2\cdot3=324,$$ $$3\cdot2\cdot3\cdot3\cdot2\cdot3=324.$$ Thus the number of $6$ letter words with the following conditions is $324+324+324=972$.
