# Probability – Permutation and Combination

Hey guys. Welcome to this video on **Probability**.

In this video we’ll take a look at two different types of probability using permutation and combination.

Now, the first thing you need to know about permutation and combination is when to use them.

When working with a problem where permutation or combination is needed, to distinguish which one, all you need to do is ask yourself the question “does order matter?”

For example, if I tell you that I made a quiche with sweet potatoes, broccoli, and bell peppers it doesn’t matter which order I say it in. I could say I made a quiche with broccoli, bell peppers, and sweet potatoes, and it wouldn’t matter. So, in this case, order is not important. However, if I tell you the password to my computer is 876, then the order of those numbers is important. If I tried 786, I would be denied access.

Now, when order is not important, like in the first example, then it is a **combination**; but, when order is important it is a **permutation**.

There is a little joke that people often make. They say, “a combination lock should really be a permutation lock.”

Alright, let’s take a look at a couple different problems. We will look at the formula for both permutation and combination, as well as how to spot whether or not order is important.

An art gallery has twelve paintings by a local artist and wants to arrange four of them on the same wall. How many ways are there to do this?

Alright, let’s look at our problem and identify whether or not order is important. Well, if an art gallery wants to arrange them a specific way then the order must be important. So, basically all we need to find is how many ways there are to do this.

Now, the formula for **permutation** is:

\(_{n}^{}\textrm{P}_{r}=\frac{n!}{(n-r)!}\)

n= total # in the set

r= how many elements are in the arrangements

So in this case, if *n* is the total number in our set, then \(n=12\text{ paintings}\), and if *r* is how many elements in the arrangements, then \(r=4\).

And this right here is all the information we need to be able to then apply it in our formula.

\(_{n}^{}\textrm{P}_{r}=_{12}^{}\textrm{P}_{4}=\frac{12!}{(12-4)!}=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\)

Now, once we write it all out, we can see that our 8 through 1 on the bottom and the top can cancel out with one another. And we’re left with \(12\cdot 11\cdot 10\cdot 9=11,880\).

Another way to think through what is going on is this:

You have four different places on a wall, so that’s 1, 2, 3, 4. Now, in the first place on our wall there are 12 different painting options to choose from. Now, once we’ve chosen one from our 12, we’re left with 11 to go in our second option, 11 different painting options. Then once we choose one from our 11 options, we’re left with 10 to go on Spot 3. And once we choose one from that, we’re left with 9 to go in our Spot 4.

So, the total permutations, the different ways that we could arrange these 4 paintings, since we care about the order in which we place them on the wall, is \(12\times 11\times 10\times 9\).

If we look closely, we can see that this \(12\times 11\times 10\times 9\), once we write it out in this way, when we think of it in terms of having 4 spots on a wall and 12 in the first, and when we take one away, we have 11, and so on, we can see that this is the exact same thing we have when we actually use our permutation formula. Once we cross out our 8 through 1 here on the top and on the bottom, we’re left with \(12\times 11\times 10\times 9\).

Now, each of these ways are equal, my goal in showing you each is just to help you get down to the foundation of where we got this permutation formula.

Let’s take a look at an example of a **combination** problem.

A person playing poker is dealt five cards. How many different hands are possible for the poker player to have been dealt.

So, for this problem we don’t care about the order in which the cards are placed in this person’s hand. Order doesn’t matter for this problem, so we know that we will need to use combination.

Here is our **combination** formula:

\(_{n}^{}\textrm{C}_{r}=\frac{n!}{r!(n-r)!}\)

n= total # of playing cards

r= cards in hand

So, since *n* is equal to our total number of playing cards, we know \(n=52\). Now, it doesn’t say it in our problem, but we are expected to know that there are 52 cards in a standard playing deck. Now *r* is the number of cards in the person’s hand at a time, so \(r=5\), and that, it tells us that right here, “dealt five cards.”

Now, we have to be careful, because the formula for permutation and combination are very similar. So let’s talk about the only difference between the two, and what that difference is representing.

The difference is that in our combination formula we have \(r!\) being multiplied by our \((n-r)!\), but we don’t have that in our permutation formula. So, what does it mean? What does this \(r!\) times everything in the denominator tell us? It tells us that we are adjusting our permutation formula by reducing it by how many ways the cards could be in order, to get our combination formula. And the reason we reduce it by how many ways the cards could be in order is because we don’t care about the order when it comes to combinations. That is the very reason we are using the combination formula, because order doesn’t matter.

Hopefully, that helps you to better understand the fundamental difference between permutation and combination.

Now, let’s plug in our number to our combination formula, and solve.

\(_{n}^{}\textrm{C}_{r}=\frac{52!}{5!(52-5)!}=\frac{52!}{5!\cdot 47!}=2,598,960\)

Now, in this example, I didn’t write out the \(52!\) all the way, but that’s because my main purpose in showing you guys how to do this problem is to teach you guys the difference between permutation and combination, and to show you how to apply your numbers into the formula.

I hope that this video has been helpful to you. For further help, be sure to check out more of our videos by subscribing to our channel below.

See you guys next time!

## Probability Practice Questions

**Question #1:**

Kim works at a smoothie bar that offers 42 different options for what fruits, juices, and add-ins customers can request in their smoothies. Each smoothie can have five unique items in it. How many possible smoothies can be made?

620,495

747,116

812,944

850,668

**Answer:**

Because order does not matter, this is a combination problem. Using \(n=42\) and \(r=5\), we use the combination formula below:

\(_nC_r=\text{ }_{42}C_5=\frac{42!}{5!(42-5)!}=\frac{42!}{5!37!}=\frac{42×41×40×39×38×37!}{5×4×3×2×1×37!}=\frac{102{,}080{,}160}{120}=850{,}668\)

**Question #2:**

Twenty individuals are signed up to compete in an upcoming triathlon where the first five places win cash prizes. How many possible ways can the top five be arranged from this pool of twenty people?

1,860,480

2,096,456

2,481,797

2,612,054

**Answer:**

From twenty competitors, we are concerned with who will make the top five. We are also interested in the order of the top five, meaning we will need to use the formula for permutations as follows:

\(_nP_r=\text{ }_{20}P_5=\frac{20!}{(20-5)!}=\frac{20!}{15!}=\frac{20×19×18×17×16×15!}{15!}=1{,}860{,}480\)

**Question #3:**

How many ways can the four numbers 1, 3, 5, and 7 be arranged? (For example, 5731, 1537, etc)

24

36

49

120

**Answer:**

Because we are dealing with ordering these numbers, we will use the formula to calculate the number of permutations where \(n=4\) and \(r=4\). Remember that the 0! in the denominator is equal to 1.

\(_4P_4=\frac{4!}{(4-4)!}=\frac{4!}{0!}=4×3×2×1=24\)

**Question #4:**

Charles is at the ice cream parlor, deciding which flavors of ice cream he wants. One bowl of ice cream comes with three scoops, and Charles wants to try three separate flavors. If there are 21 flavors to choose from, how many possible bowl combinations are there?

880

1,330

1,520

2,190

**Answer:**

From 21 flavors, Charles will choose 3. So we use the formula for combinations (because order doesn’t matter) where \(n=21\) and \(r=3\).

\(_{21}C_3=\frac{21!}{3!(21-3)!}=\frac{21×20×19×18!}{3!×18!}=\frac{21×20×19}{3×2×1}=1{,}330\)

**Question #5:**

Suppose that Charles chooses to get an ice cream cone instead of a bowl, where the three scoops are stacked one on top of another. How many possible ways could he arrange three flavors stacked? Remember, there are 21 flavors to choose from.

5,140

5,960

6,220

7,980

**Answer:**

In this problem, order matters. So we use the formula for permutation, where \(n=21\) and \(r=3\) once again.

\(_{21}P_3=\frac{21!}{(21-3)!}=\frac{21!}{18!}=\frac{21×20×19×18!}{18!}=21×20×19=7{,}980\)