What's the difference between a combination and a permutation? Let's contrast two problems:
Think about the difference between two problems:
1) How many ways can we make a team of 3 people from 8 people?
2) How many ways can we make a team consisting of a President, VP, and Treasurer from 8 people?
In both of these problems, we are choosing 3 people out of a group of 8 people. In problem (1), there is only one possible combination with each group of 3 people. The positions on the team are the same. For example, I choose Ann, Bob, Cal. Choosing Ann, Bob, Cal is the same as choosing Bob, Cal, Ann, or Cal, Ann, Bob, etc. There are actually 3! = 6 different ordering of these three people.
But for (1), all these orderings don't matter -- they all represent the same ONE combination of Ann, Bob, and Cal.
Another way to think about it: We choose three people and put them together in a room. We don't care about their roles, their seating, or what they are doing. We only care who is in the room.
For that reason, (1) is a combination problem.
BUT IN PROBLEM (2) we are not only choosing 3 people, but we are giving them 3 different positions:
Pres, VP, and Treasurer.
Now, even with the same combination of 3 people, there are 3! ways to assign these people the different positions:
For example, choosing
Ann = Pres, Bob = VP, and Cal = Treasurer is DIFFERENT than
Cal = Pres, Ann = VP, and Bob = Treasurer.
So our answer to (2) is 3! times the answer to (1). It's 8P3 = 8!/5! This is equivalent to using the Fundamental Counting Principle: 8 * 7 * 6. 8 choices for President, then 7 choices for VP, then 6 choices for Treasurer.
Therefore, in (2), both which 3 people we choose and also the order in which we arrange them matters, which means that (1) is a Permutation/FCP problem.
Equivalent to assigning different positions could be assigning the people to different seats, or cities, etc.
To go back to the "people in a room example," we now no longer care only about who is in the room. We care about how they are arranged in the room, or in what order they went into the room.
If we only care about what things we choose, then we only care about the combination.
If the order/position/role of the things we are choosing are distinct, then we have a permutation.
Now, remember, the permutation formula is equivalent to multiplying the choices for each stage using the Fundamental Counting Principle. That's why there is no need to ever use the permutations "formula:"
For example, 8P3 above is equivalent to 8 * 7 * 6.
nPr = n!/(n-r)!
8P3 = 8!/(5!)
When we evaluate this, 8! = 8 * 7 * 6 * (5 * 4 * 3 * 2 * 1) and the (5 * 4 * 3 * 2 * 1) cancels out with the 5! in the denominator. So what do you we end up with?
8 * 7 * 6
In the same way, the combinations formula is the same thing as starting with the FCP and then dividing by the number of positions you have.
nCr = n!/((n-r)!r!) <--- notice this is the same as nPr except with an extra r! at the bottom! This makes sense, because there are r! ways to arrange the r things we choose. Since we want to count these different arrangements of r things as the same, simply using a permutations calculation gives us r! too many things.
So we just need to take the permutations calculation and divide by r!.
For 8C3, then, take 8 * 7 * 6 and divide by 3!. That's it!
Here's one more example:
Suppose we have 9 distinct balls and you are choosing 4 to drop into 4 distinct boxes.
9 * 8 * 7 * 6
Now, if you're dropping them all into one box and you only care about which balls we chose, then then we divide by 4!, which is the number of ways to arrange 4 things.
So we end up with 9 * 8 * 7 * 6/ 4!
Which is the same as 9C4.
Trickier problems will involve more than just an outright calculation like those above. But if you understand the above well, you have a great foundation to work with :) Here's a GMAT post with some practice :)