2.3.1: The Multinomial Part 1
With the topic of permutations under our belt, we will now consider the following question.
Suppose we have two identical copies of Homer's Odyssey and one copy of Plato's Republic. How many ways are there to arrange these three books on a bookshelf?
Critique This Solution:
We have three objects and by our theorem from the last section, there are \( 3! = 3 \times 2 \times 1 = 6 \) different ways to arrange 3 objects. Is this correct? Why or why not?
- Answer
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The above answer is incorrect! If we were given three distinct objects, then there are certainly 3! ways to arrange 3 distinct objects. However, we do not have three distinct (or unique) objects in our question. Notice that some of the objects are identical to each other.
In order to answer the above question, allow us to enumerate all the possibilities. We will denote each copy of Homer's Odyssey by H and we will denote the single copy of Plato's Republic by P. Enumerating yields the following:
- \( (H, H, P) \)
- \( (H, P, H) \)
- \( (P, H, H) \)
We see that there are only three possible arrangements.
The point of the above example is to illustrate that as soon as we break one of our assumptions in the last section (namely that we have distinct elements), our entire theory seems to fall apart. The goal of this mini-section is to develop a formula which tells us how many ways we can arrange \(n\) objects when some of the objects are identical to each other.
In order to answer the above question, we first make a definition.
Objects are said to be of the same type if the objects are identical/indistinguishable from each other.
For instance, in the opening exercise, the \(H\) and \(H\) are of the same type since each copy of Homer's Odyssey is identical to each other. We now present the answer to our question in the form of a theorem. The proof of the theorem is discussed at the end of this section.
Given \(n\) objects where:
- \(n_1\) objects are of Type 1,
- \(n_2\) objects are of Type 2,
- \( \ldots \)
- \(n_k\) objects are of Type k,
and \( n_1 + n_2 + \ldots + n_k = n \), the number of different ways to arrange the \(n\) objects is given by
\[ \displaystyle \frac{n!}{n_1 ! n_2 ! \ldots n_k !} \nonumber\ \]
Suppose we have two identical copies of Homer's Odyssey and one copy of Plato's Republic. How many ways are there to arrange these three books on a bookshelf?
- Answer
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In this problem, we have two objects of Type 1 (where Type 1 is Homer's Odyssey ) and one object of Type 2 (where Type 2 is Plato's Republic ). By our theorem , there are \[ \displaystyle \frac{3!}{2 ! 1! } = 3 \nonumber\ \] different arrangements.
How many different arrangements are there of the word AAABB?
- Answer
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In this problem, we have three objects of Type 1 (where Type 1 are the A's) and two objects of Type 2 (where Type 2 are the B's). By our theorem , there are \[ \displaystyle \frac{5!}{3 ! 2! } = 10 \nonumber\ \] different arrangements.
Allow us to explore some additional questions related to the example above because it may be helpful in understanding the the proof of the theorem .
- List of all arrangements of the word AAABB.
- Suppose the objects of Type 1 were distinct. That is, assume all of the A's are distinct (we can achieve this by labeling or indexing the A's). How many different arrangements are there now?
- Answer
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Here are all of the arrangements of AAABB:
(A,A,A,B,B)
(A,A,B,A,B)
(A,B,A,A,B)
(B,A,A,A,B)
(B,A,A,B,A)
(B,A,B,A,A)
(B,B,A,A,A)
(A,A,B,B,A)
(A,B,B,A,A)
(A,B,A,B,A)
If the A's were all distinct, then we have to arrange the following: \( A_1, A_2, A_3, B, B \). For convenience, we will label them as (A1, A2, A3, B, B). We can do this neatly via a grid:
A1, A2, A3 A1, A3, A2 A2, A1, A3 A2, A3, A1 A3, A1, A2 A3, A2, A1 (A,A,A,B,B) (A1,A2,A3,B,B) (A1,A3,A2,B,B) (A2,A1,A3,B,B) (A2,A3,A1,B,B) (A3,A1,A2,B,B) (A3,A2,A1,B,B) (A,A,B,A,B) (A1,A2,B,A3,B) (A1, A3,B,A2,B) (A2,A1,B,A3,B) (A2,A3,B,A1,B) (A3,A1,B,A2,B) (A3,A2,B,A1,B) (A,B,A,A,B) (A1,B,A2,A3,B) (A1,B,A3,A2,B) (A2,B,A1,A3,B) (A2,B,A3,A1,B) (A3,B,A1,A2,B) (A3,B,A2,A1,B) (B,A,A,A,B) (B,A1,A2,A3,B) (B,A1,A3,A2,B) (B,A2,A1,A3,B) (B,A2,A3,A1,B) (B,A3,A1,A2,B) (B,A3,A2,A1,B) (B,A,A,B,A) (B,A1,A2,B,A3) (B,A1,A3,B,A2) (B,A2,A1,B,A3) (B,A2,A3,B,A1) (B,A3,A1,B,A2) (B,A3,A2,B,A1) (B,A,B,A,A) (B,A1,B,A2,A3) (B,A1,B,A3,A2) (B,A2,B,A1,A3) (B,A2,B,A3,A1) (B,A3,B,A1,A2) (B,A3,B,A2,A1) (B,B,A,A,A) (B,B,A1,A2,A3) (B,B,A1,A3,A2) (B,B,A2,A1,A3) (B,B,A2,A3,A1) (B,B,A3,A1,A2) (B,B,A3,A2,A1) (A,A,B,B,A)
(A1,A2,B,B,A3) (A1,A3,B,B,A2) (A2,A1,B,B,A3) (A2,A3,B,B,A1)
(A3,A1,B,B,A2) (A3,A2,B,B,A1) (A,B,B,A,A) (A1,B,B,A2,A3) (A1,B,B,A3,A2) (A2,B,B,A1,A3) (A2,B,B,A3,A1) (A3,B,B,A1,A2) (A3,B,B,A2,A1) (A,B,A,B,A) (A1,B,A2,B,A3) (A1,B,A3,B,A2) (A2,B,A1,B,A3) (A2,B,A3,B,A1) (A3,B,A1,B,A2) (A3,B,A2,B,A1) We see that there are a total of 60 different arrangements of the word \( A_1, A_2, A_3, B, B \). The 60 stems from the fact that there are initially 10 arrangements of the word when the A's are identical. Once the A's are no longer identical, there are \(3!\) ways to permute them and so there are a total of \( 10 \times 3! = 60 \) different arrangements.
In the Olympics, there are 10 competitors. Of the 10 competitors, 4 competitors represent the United States, 3 competitors represent China, 2 competitors represent England and 1 competitor represents Canada. If the results just lists out the countries of the players in the order that they placed, then how many outcomes are possible?
- Answer
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According to the question, an outcome is a ranking of the countries. That is, an outcome looks like \( ( \text{ United States, England, Canada, England, China, China, China, United States, United States, United States} ) \). We see that we have four objects of Type 1 (the competitors representing The United States), 3 objects of Type 2 (the competitors representing China), 2 objects of Type 3 (the competitors representing England), and 1 objects of Type 4 (the competitor representing Canada). By our theorem , there are \[ \displaystyle \frac{10!}{4! 3! 2 ! 1! } = 12,600 \nonumber\ \] different rankings.
We end this section with a seemingly unimportant result followed by the proof of the theorem .
Suppose we have \(n\) elements where \(k\) elements of Type 1 and \(n-k\) elements are of Type 2.
How many different ways can we arrange the \(n\) elements?
- Answer
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By our theorem , there are \[ \displaystyle \frac{n!}{k! (n-k)! } \nonumber\ \] different arrangements. We may see this expression again in the next section!
Given \(n\) objects where:
- \(n_1\) objects are of Type 1,
- \(n_2\) objects are of Type 2,
- \( \ldots \)
- \(n_k\) objects are of Type k,
and \( n_1 + n_2 + \ldots + n_k = n \), the number of different ways to arrange the \(n\) objects is given by
\[ \displaystyle \frac{n!}{n_1 ! n_2 ! \ldots n_k !} \nonumber\ \]
- Proof
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Let \( d \) denote the number of distinct arrangements of the \( n\) objects. We wish to show that \( d = \frac{n!}{n_1! n_2! n_3! \ldots n_k!} \).
Notice that if the objects of Type 1 were distinct (which can be achieved by labeling or making an index), then there are \( d \times n_1! \) different arrangements.
Also, if the objects of Type 2 were distinct (which can be achieved by labeling or making an index), then there are \( d \times n_1 ! \times n_2! \) different arrangements.
Also, if the objects of Type 3 were distinct (which can be achieved by labeling or making an index), then there are \( d \times n_1! \times n_2! \times n_3! \) different arrangements.
\( \ldots \)
Finally, if the objects of Type \( k \) were distinct, (which can be achieved by labeling or making an index), then there are \(d \times n_1! \times n_2! \times n_3! \times \ldots \times n_k! \) different arrangements.
At this point in time, we have indexed a total of \(n_1 + n_2 + n_3 + \ldots + n_k \) objects. By assumption, \(n_1 + n_2 + n_3 + \ldots + n_k \ = n \) and so at this point in time, all \(n\) objects are distinct.
Thus, we have counted the number of different arrangements of these \(n\) distinct objects in two different ways. One way is \(d \times n_1! \times n_2! \times n_3! \times \ldots \times n_k! \) and the other way is \(n!\).
Since these expressions count the same quantity, they are equal to each other. That is, \(d \times n_1! \times n_2! \times n_3! \times \ldots \times n_k! = n! \). Solving for \(d\) yields
\[ \displaystyle d = \frac{n!}{n_1 ! n_2 ! \ldots n_k !} \nonumber\ \]
which is precisely what we needed to show.