I'm studying beginner combinatorics. How do I know which one to use? I'm studying combinations and permutations in high school. How do I know exactly when to use which? On tests, my teacher expects me to know when to use permutations and when to use combinations. I've read all over that if "order matters", use permutations, and if "order doesn't matter", use combinations, but that seems vague and prone to error.
I've done questions where it seems as though combinations are to be used since order doesn't matter, but permutations are to be used instead. I just don't understand when to use which.
Could someone explain when to use which and how to develop intuition for these problems? I get that I should just do them and develop an intuition, but I will be given unusual, never seen before problems on tests and I don't want a 50/50 chance of solving the problem.
Thanks.
 A: Ask yourself, are you assigning items to different positions or places? If yes, then order matters. You have a permutation. I'll try to explain through a few examples. Hopefully the level of the problems is appropriate.

There are 15 contestants in a race. In how many ways can we award gold, silver, and bronze medals? 

In this case, we are assigning people to different positions represented by the medals. First is clearly more advantageous than second. It's a permutation: $15\ P\ 3$.

There are 40 contestants in a raffle. Six will win tickets to Disney Land. How many ways can we select the winners?

Are we assigning people to positions? Not really. The first person picked and the second person picked both get the same reward. Everyone who is chosen is a winner. If we conduct a thought experiment and switch the order of selection, nothing changes. They have the same position, so to speak. Combination $40\ C\ 6$.

You and four friends are about to sit in five consecutive seats at a movie theater. How many arrangements are possible?

We are literally assigning people to different physical positions - the seats.  Consider how different the movie may feel for you if you had an aisle versus the center seat. The order matters! Permutation: $5\ P\ 5$. 

Sometimes a thought experiment will help you determine whether order matters. In the next example, we purposefully consider altering the order of selection to see if the result changes.

How many line segments can I draw connecting six distinct points in a plane, no three of which are collinear?

First think, how many points do we choose to create a line segment? Two points. Okay, now imagine a line segment made by two of the points. First pick $A$, then pick $B$. We have segment $\bar{AB}$. Now deliberately consider changing the order in which you picked the points: select point $B$ then point $A$. You have segment $\bar{BA}$. But that's the same as $\bar{AB}$. The order did not matter. Combination $6\ C\ 2$.

I noticed your example question in the comments. It's above the level of the above problems, but let's try to apply the same ideas.

In her repertoire, Esther has seven songs from which she will perform two at each school concert. In how many ways can the programs for three consecutive shows be arranged so that she does not perform the same song two shows in a row?

I'm interpreting this question as stating that each night's program can be constructed without caring about which song comes first or second. Whether we play $AB$ on the first night or $BA$ doesn't matter, as we only care to differ the songs from the previous night. If we just care about what two songs can be paired up every night, we have a combination. 
For first night we have $7\ C\ 2 = 21$ possible programs. We're barred from playing the first two songs on the second night, so we next have $5\ C\ 2 = 10$ programs that can be created. For the third night, there are again five songs that differ from night two, so $5\ C\ 2 = 10$.
Then the answer is $21 \cdot 10 \cdot 10 = 2100$.
A: Use combinations when order doesn't matter. Use permutations when order matters.
Alternatively derive them on the spot.

Rule of product:
A decision with $m$ outcomes and another with $n$ outcomes regardless of which option was taken for the first has $mn$ outcomes.


How many ways are there to arrange 4 things chosen from a set of 10?

For the first thing we have 10 options, for the second we have 9 left, for the third we have 8, for the 4th we have 7, so the answer is:
$10*9*8*7=\frac{10!}{6!}=\ ^{10}P_4$


How many ways are there to choose 4 things from a set of 10.

There are $\frac{10!}{6!}$ ways to arrange them. But say we're given 4 things, there are 4 ways to choose the first, 3 for the second... so that means given 4 things there are $4*3*2*1$ ways to actually arrange them.
So each of our $\frac{10!}{6!}$ ways will be repeated $4!$  times (as these are all distinct arrangements) dividing by this we get: $\frac{10!}{6!4!}$ which you'll note is $\ ^{10}C_4$

