Confused about when to use permutations or combinations How many baseball teams can be formed from 15 players if 3 only pitch and the others play any of the remaining 8 positions?
I'm thinking that this is permutations, but my teacher says it is combinations. 
However, there is this problem:
Five people have applied for three different positions in a store. If each person is qualified for each position, how many ways can the positions be filled?
I know that this problem uses permutations for sure, I have already check the answer, and I notice that this is very similar wording to the baseball problem. Why is the baseball problem using combinations and not permutations then? Isn't it using permutations because each player is unique and different, and therefore, specific order matters? Very confused about this.
Thanks for any help!
 A: For the baseball problem,
you just have to choose a pitcher from the $3$ eligible, and a team from the other $12$, thus $\binom31\binom{12}8$
There is no indication that other than the pitcher, players have to be allotted distinct positions.
I hope you can see why it is a combination problem
In the other problem, the posts (positions) are distinct, hence a permutation problem.
A: Most "combination" problems use permutations and many "permutation" problems use combinations.   Except (and not even always then) in the simplest of cases, it is not often exclusively one or the other; it is usually only "mostly" or "from a certain point of view". 
Rather than trying to classify a problem as one or the other, simply ask what are you counting, and how best to do it.

How many baseball teams can be formed from 15 players if 3 only pitch and the others play any of the remaining 8 positions?

Count the ways to select three from fifteen items and then eight from the remaining.   (Combinations!)  $$\binom{15}{3}\binom{12}{8}~=~225225$$
Count the ways to arrange the fifteen players, divide by ways to arrange (a) the first three (pitchers), (b) the next eight, (c) the last four (non-players).   (Permutations!) $$\dfrac{15!}{3!~8!~4!}~=~225225$$ 
But wait, these things are equal!

Five people have applied for three different positions in a store.   If each person is qualified for each position, how many ways can the positions be filled?

Likewise we can count ways to select one person for each position (combinations), we can count ways to select three people and arrange them among the positions (permutations), or we can count ways to arrange the five players and divide by ways to arrange the two who don't get positions (permutations again!).
$$\binom{5}{1}\binom{4}{1}\binom{3}{1} ~=~ \binom{5}{3}3!~=~ \dfrac{5!}{2!}~=~60$$
