How many ways can a slate of 4 (distinct) officers from 20 people, etc. In how many ways can a slate of 4 officers (president, vice-president, secretary, treasurer) be selected from among 20 people? 
$$4! {20 \choose 4}$$
In how many ways can a committee of three be selected from among 10 people?
$$3! {10 \choose 3}$$
I wanna make sure I'm doing the questions above correctly, any feedback is greatly appreciated 
 A: Your first answer is correct, and as you accounted for, the order of the four people selected to be officers matters.  We have 4 distinct officer positions to fill, and there are $4!\dbinom{20}{4} = (20\cdot 19\cdot 18\cdot 17)$ ways to do that.

Your second answer is incorrect:  The number of ways a committee of three can be selected from ten people is simply $$\binom{10}{3} = \dfrac{10!}{7!3!} = \frac{10\cdot 9\cdot 8}{6} = 120$$  
If the three chosen consist of person A, person B, and person C, any permutation of those three people will not change the committee.  If we represent the three chosen from ten as $(A, B, C)$, then order does not matter: if you're in the committee you are in the committee, period. So $$(A,B,C) = (A,C,B) =(B,A,C) = (B,C,A) = (C,A,B) = (C, B, A).$$
A: You already have the answer in the Comment of @ThomasAndrews. This is just
to show some terminology and notation. While I was typing this another very nice
explanation from @amWhy appeared (+1). Please leave a Comment to say whether you
now understand the distinction between the two problems, or Accept one of the
answers to indicate the same thing.

The first problem is to choose an ordered sample: Pres, VP, Sec, Treas.
Some books call this a permutation problem.
Ways to choose Pres: 20
Then ways to choose VP: 19
So answer is 
$${}_{20}P_4 = 20\cdot 19\cdot 18\cdot 17 = 20!/16! = 4!\frac{20!}{4!\cdot 16!}
= 4!{20 \choose 4} = 4!({}_{20}C_4).$$
The second problem is to choose an unordered sample.
Some books call this a combination problem: ${}_{20}C_3 = {20 \choose 3}.$
