Factorials with Identical Items

Background

We talked about Factorials when we have all the items being distinguishable (we can tell each item apart from each other item). But what happens when we have duplicate items/indistinguishable items?

Question

Stephanie has six special teacups that she wants to display in a row on a shelf. In how many ways can she display them if they:

1. are all distinguishable (are all different),
2. are all indistinguishable (are all the same),
3. five are red and one is blue
4. four are red and two are blue
5. four are red, one is blue, and one is yellow
6. two are red, two are blue, two are yellow 

Answer

1. 
2. 
3. 
4.  
5.  
6.  

Analysis

Before we move into areas dealing with identical items, let's first work through a problem we worked through already in the Factorials entry:

Question 1

Where there are six things that are all different and we're putting them in a row, we can put all of the 6 items into the first place.

We can then take one of the remaining 5 items and place it in the second position.

We can then take one of the remaining 4 items and place it in the third position.

And so on. And so we end up with:

6 x 5 x 4 x 3 x 2 x 1 = 6! = 720

Question 2

Now let's work through a question where all the items are the same. Let's say that each item is labeled R for Red. Let's lay them out:

RRRRRR

It doesn't matter how I shuffle the teacups - it's all the same. Therefore, there is only 1 way to arrange them.

Let's now get to this answer mathematically.

Let's start looking at this by looking at 3 items. If they are all different, we have 3! = 6 ways to arrange the items:

ABC
ACB
BAC
BCA
CAB
CBA

but if B and C are exactly the same so that we have 2 B's, we have:

ABB
BAB
BBA

Let's look at ABB. There are two arrangements in here - ABC and ACB - but because B and C are identical, we don't know which is which.

The same is true of BAB. And also true of BBA.

The way we can express this mathematically is that we start with the total number of items to arrange: 3 in our current example, and take the factorial. Here we have 3!

We then divide by the number of ways we can arrange the different items. For items that are distinguishable, like in Question 1, we can divide by 1! For items that are the indistinguishable, we divide by the number in the indistinguishable items. So for two B's, we divide by 2!. And so we end up with:



We can now work out the question in the original question. There are 6 items we are arranging and there are 6 items that are identical. And so we get:



Question 3

Now we have 5 Red cups and 1 Blue cup. We can work through the arrangements by listing them out:

RRRRRB
RRRRBR
RRRBRR
RRBRRR
RBRRRR
BRRRRR

And we can also work out the problem mathematically:



Question 4

We can list out the different arrangements, but let's do this mathematically first:



RRRRBB
RRRBRB
RRBRRB
RBRRRB
BRRRRB

RRRBBR
RRBRBR
RBRRBR
BRRRBR

RRBBRR
RBRBRR
BRRBRR

RBBRRR
BRBRRR

BBRRRR

Question 5

The number of arrangements is becoming enough that space becomes short. Let's just work this out mathematically. There are 4 Red teacups, 1 Blue, and 1 Yellow. Therefore, we take as the numerator 6! and we divide by 4! (the number of ways to arrange the 4 Red teacups) and 1! each for the 1 Blue and 1 Yellow teacups:



Question 6

We now have three groups of two colours each. We therefore have 6! in the numerator and then we divide by 2! for each of the three groups (which we can express in exponential form):



Vocabulary used:

For more information check out these links (comment to add your favourite link):

Where might you have come from?

Fact-orials Index

Combinatorics:

• Factorials

Where might we go?

Factorials in a Circle

Background

We know how to calculate the number of ways we can arrange distinct items (such as books each with different title) in a row (like on a shelf). Does anything change if we arrange those same items around a table?

Question

Adam, Barb, Carol, and Derek are going to play cards at a round table. The players don't care about which chair they sit in, but they do care about what order the people are around them. In how many different ways can the players sit?

Answer

(n-1)! = (4-1)! = 3! = 6

Analysis

Let's start with looking at the players, ABCD, can stand in a row (perhaps they are waiting outside the venue in order to play cards). From the Factorials entry, we know that there are 4! = 24 ways for them to stand in a row.

Now they approach the table they will play at. A will sit down first. Does it matter which chair he sits in? No - they are all equal. Where A sits has no bearing on anything.

B will now sit and gets to decide she'll sit on A's right, left, or straight across. So B has three choices as to where to sit.

C will then sit and can pick from the remaining two spots.

And then D sits in the last remaining spot.

And so the number of possible ways for the four people to sit is 3 x 2 x 1 = 6.

Remember that when we multiply natural numbers up to a given number, that's a factorial calculation that we can symbolize using an exclamation symbol. So in this case we have:

3! = 3 x 2 x 1 = 6

And in general, when we seat people at a round table, the number of ways we can seat them, looking only at the relationships of the people to each other, we can say that the number of ways they can be arranged is:

(n-1)!

Vocabulary used:

For more information check out these links (comment to add your favourite link):

Where might you have come from?

Fact-orials Index

Combinatorics:

• Factorials

Where might we go?

Factorials

Background

There are many different ways multiplication can be used. One way is to look at the numbers of ways things can be arranged...

Question

In how many ways can 3 distinguishable books be arranged in a row on a shelf? How about 4 books?

Answer

3! = 6 ways

4! = 24 ways 

Analysis

Let's start looking at this topic by looking at the three books. Let's say they are a book each on Accounting, Biology, and Chemistry (and so we have books A, B, and C). How can we put them on the shelf?

Let's take book A and put it on the shelf first. I can now choose to put either book B or book C next to it. Let's do book B, which means that book C has to be the last book I put up. That gives the arrangement on the shelf of:

ABC

Now, I could have chosen to put book C next to A and get this arrangement:

ACB

That's 2 arrangements starting with A. I could have started with book B and gotten these two options:

BAC
BCA

and the same with starting with book C:

CAB
CBA

That's 6 choices in total for how I can arrange the books.

Let's take a closer look to see what's going on.

For each starting book - like A - I had two ways to arrange the other two books. And I can start with three different books. And so I can say that the number of ways to arrange the books is:

3 x 2 = 6

I'm going to go one step further and include a x 1 in there - it won't change the outcome but it will help in more complicated problems:

3 x 2 x 1 = 6

When I have a problem where I am putting distinguishable objects (objects I can tell apart) and I am putting them into a distinguishable order (such as putting them into a row), I can use this type of math to find the number of ways I can order them.

This is called a "factorial" (and yes, this is where the name for this resource came from!)

In fact, this type of math is so common that we have notation to show that we'll using it. We use an exclamation mark to show that we're going to multiply natural numbers up to the number indicated. In this case, we can say that:

3 x 2 x 1 = 3! = 6

In general, we use the notation of n! for factorial calculations.

So now let's look at what happens if we add a 4th book to the mix, book D. Let's put D on the shelf first - now see that there are 6 ways to arrange books A, B, and C (what we just found above). Now let's put A in front - there are 6 ways to arrange books B, C, and D. And we can do the same with B and C as the first book. So in total, we have:

4 x 6 = 4 x 3! = 4 x 3 x 2 x 1 = 4! = 24

Vocabulary used:

For more information check out these links (comment to add your favourite link):

Where might you have come from?

Fact-orials Index

Operations:

• Multiplication

Where might we go?

Combinatorics:

• Factorials in a Circle
• Factorials with Identical Items
• Permutations