Factorials Examples

Maisy is working the counter at Shmaskin Robbins. A hungry customer orders a triple scoop ice cream cone with strawberry, chocolate, and vanilla ice cream.

How many different ways could she stack the ice cream flavors on top of each other?

You could answer the question by listing all of the possible orders of the ice cream (top to bottom).

1. Chocolate-Strawberry-Vanilla
2. Chocolate-Vanilla-Strawberry
3. Strawberry-Vanilla-Chocolate 
4. Strawberry-Chocolate-Vanilla
5. Vanilla-Strawberry-Chocolate
6. Vanilla-Chocolate-Strawberry

Or, use factorials to get the answer much faster: 

3! = 3 × 2 × 1 = 6

Now, what's the probability that chocolate will be on top?

Two of these combinations have chocolate on top 

1. Chocolate-Vanilla-Strawberry 
2. Chocolate-Strawberry-Vanilla

So the probability is 2 out of the 6 possible combinations:

Tom, Tristan, Luca, and Pablo are lined up and ready to be picked for kickball teams.  How many different ways can they be picked from first to last?

We could list the ways, but that isn't necessary now that we know about awesome factorials. 

The number of ways to pick the four boys is 4! = 4 × 3 × 2 × 1 = 24.

If all boys were picked in a random order, what is the probability that Tom will luck out and be picked first?

To answer the probability question we need to look at all the ways that Tom could be picked first. Here they are:

1. Tom-Tristan-Luca-Pablo 
2. Tom-Tristan-Pablo-Luca
3. Tom-Pablo-Luca-Tristan
4. Tom-Pablo-Tristan-Luca
5. Tom-Luca-Pablo-Tristan 
6. Tom-Luca-Tristan-Pablo

If we fix Tom as the kid picked first, and arrange the other three, we find 6 different ways, which is 3!.

3! = 3 × 2 × 1 = 6 

So the probability that Tom is picked first is 6 out of a possible 24 combinations. The probability is:

You have 9 antique glass vases and want to display 3 on a shelf. How many different ways are there to do this?

Nothin’ like setting them all out and seeing which combo looks best. The first vase is made of an old dark blue glass. If we put that up first, we can either do vase 2, 3, 4, 5, 6, 7, 8, or 9 next.

If we put vase 1 first and vase 2 in the middle, that leaves vase 3, 4, 5, 6, 7, 8, or 9 for the last spot.
If we put vase 1 first and vase 3 in the middle, that leaves vase 2, 4, 5, 6, 7, 8, or 9 for the last spot.
If we put vase 1 first and vase 4 in the middle, that leaves vase 2, 3, 5, 6, 7, 8, or 9 for the last spot.
Writing this out in a list, we get:

It appears that putting vase 1 first leaves 8 choices for the middle vase and 7 choices for the last vase. So that’s 1 × 8 × 7. But that’s just if we put vase 1 first.

We have that same number of options for each vase that we put in first place. Since there are 9 vases, it seems we have 9 × 8 × 7 different ways to arrange 9 vases in 3 spots. Here are the first few possible outcomes:

And so on.

Since 9 × 8 = 72 and 72 ×7 = 504, we have 504 choices. Trying them all out? Not a good idea. How about we just buy two more shelves and display them all? Hmmmm, how many different ways could we arrange the 9 vases if we did that? That’s a little light calculating to keep us up tonight.