**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 *t**hen 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 :)

## Comments