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?