5.2: Permutations and Combinations
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Section 1: Factorial Notation
\(n!\) denotes the product of natural numbers from \(1\) to \(n\), i.e.
\(n!=n\cdot(n-1)\cdot\dots\cdot3\cdot2\cdot1\)
\(4!=4\cdot3\cdot2\cdot1=24\)
\(5!=5\cdot4\cdot3\cdot2\cdot1=120\)
\(8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=40320\)
Note that
\(0!=1\)
In combinatorics, we would want to be able to simplify the following expressions:
\(\frac{7!}{3!}\), \(\frac{7!}{4!}\), \(\frac{7!}{3!4!}\)
The easiest way to simplify the factorials is to write them out using the definition, and then reduce by the greatest common factor:
\(\frac{7!}{3!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}=7\cdot6\cdot5\cdot4\)
\(\frac{7!}{4!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}=7\cdot6\cdot5\)
\(\frac{7!}{3!4!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot4\cdot3\cdot2\cdot1}=7\cdot5\)
Frequently we will have to work with expressions like the very last one. Note that after writing out the factorials and after reducing by the common factorial we still have some common factors left which we need to reduce.
Section 2: Permutations
An ordered arrangement of distinct objects is called a permutation.
THR is a permutation of length three from letters \(\{T, H, R\}\).
HRT is a permutation of length three from letters \(\{T, H, R\}\).
HR is a permutation of length 2 from letters \(\{T, H, R\}\).
123 is a permutation of length three from letters \(\{1, 2, 3\}\).
231 is a permutation of length three from letters \(\{1, 2, 3\}\).
31 is a permutation of length 2 from letters \(\{1, 2, 3\}\).
How many permutations of length 3 out of 3 letters are there?
To create a permutation, we have to consecutively choose 3 letters. Since the letters are all distinct the number of choices at each step is decreasing by one. So by the basic counting principle the total number of permutations is \(3\cdot2\cdot1=3!\).
How many permutations of length 4 out of 4 letters are there?
To create a permutation, we have to consecutively choose 4 letters. Since the letters are all distinct the number of choices at each step is decreasing by one. So by the basic counting principle the total number of permutations is \(4\cdot3\cdot2\cdot1=4!\).
How many permutations of length \(n\) out of \(n\) letters are there?
To create a permutation, we have to consecutively choose \(n\) letters. Since the letters are all distinct the number of choices at each step is decreasing by one. So by the basic counting principle the total number of permutations is \(n!\).
How many permutations of length 2 out of 5 letters are there?
To create a permutation, we have to consecutively choose 2 letters. Since the letters are all distinct the number of choices at each step is decreasing by one. So by the basic counting principle the total number of permutations is \(5\cdot4\).
How many permutations of length 3 out of 5 letters are there?
To create a permutation, we have to consecutively choose 3 letters. Since the letters are all distinct the number of choices at each step is decreasing by one. So by the basic counting principle the total number of permutations is \(5\cdot4\cdot3\).
How many permutations of length \(k\) out of \(n\) letters are there?
To create a permutation, we have to consecutively choose \(k\) letters. Since the letters are all distinct the number of choices at each step is decreasing by one. So by the basic counting principle the total number of permutations is given by the following product:
\(n\cdot(n-1)\cdots(n-k+1)\)
In general, we want to find out how many permutations of length \(k\) out of \(n\) letters are there? Let’s denote this quantity as \(P_k^n\). Based on the observed pattern, we obtain the following formula called the permutation rule:
\(P_k^n=\frac{n!}{(n-k)!}\)
Let’s do a few applications of it.
How many permutations of length 2 out of 5 letters are there?
Solution
The answer is \(P_2^5\) which we can now compute using the formula:
\(P_2^5=\frac{5!}{(5-2)!}=\frac{5!}{3!}=5\cdot4=20\)
How many permutations of length 4 out of 10 letters are there?
Solution
The answer is \(P_4^{10}\) which we can now compute using the formula:
\(P_4^{10}=\frac{10!}{(10-4)!}=\frac{10!}{6!}=10\cdot9\cdot8\cdot7=5040\)
If the length of a permutation is the same as the number of available letters, then we replace \(k\) with \(n\) in the permutation formula and obtain the special permutation rule:
\(P_n^n=\frac{n!}{(n-n)!}=\frac{n!}{0!}=n!\)
Let’s do a few applications of it.
How many permutations of length 4 out of 4 letters are there?
Solution
\(P_4^4=4!=4\cdot3\cdot2\cdot1=24\)
How many permutations of length 6 out of 6 letters are there?
Solution
\(P_6^6=6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720\)
The following questions:
• How many ways are there to select 3 people out of 6 for a committee with distinct roles?
• How many ways are there to select 3 pieces of art out 6 for an exhibition in which the order matters?
and many other questions, all have the same answer:
\(P_3^6\)
The following questions:
• How many ways are there to select 5 people out of 9 for a committee with distinct roles?
• How many ways are there to select 5 pieces of art out 9 for an exhibition in which the order matters?
and many other questions, all have the same answer:
\(P_5^9\)
Permutations is one of the building blocks of the combinatorics.
Section 3: Combinations
A collection of distinct objects is called a combination.
\(\{T, H, R\}\) is a combination of length three from letters \(\{T, H, R\}\).
\(\{H, R, T\}\) is a combination of length three from letters \(\{T, H, R\}\).
\(\{T, H, R\}\) and \(\{H, R, T\}\) are the same combinations.
\(\{1, 2\}\) is a combination of length two from letters \(\{1, 2, 3\}\).
\(\{2, 3, 1\}\) is a combination of length three from letters \(\{1, 2, 3, 4\}\).
\(\{1, 3\}\) is a permutation of length two from letters \(\{1, 2, 3, 4\}\).
Let’s discuss the difference between permutations and combinations.
|
Permutations |
Combinations |
|---|---|
|
|
Since every combination of length \(k\) can be further arranged in order, the number of permutations of length \(k\) is \(k!\) times larger than the number of combinations of length \(k\).
In general, we want to find out how many combinations of length \(k\) out of \(n\) letters are there? Let’s denote this quantity as \(C_k^n\). Based on the fact that the number of permutations of length \(k\) is \(k!\) times larger than the number of combinations of length \(k\) we obtain the following formula called the combinations rule:
\(C_k^n=\frac{n!}{(n-k)!k!}\)
Let’s do a few applications of it.
How many combinations of length 2 out of 5 letters are there?
Solution
The answer is \(C_2^5\) which we can now compute using the formula:
\(C_2^5=\frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=\frac{5\cdot4}{2\cdot1}=10\)
How many combinations of length 4 out of 10 letters are there?
Solution
The answer is \(C_4^{10}\) which we can now compute using the formula:
\(C_4^{10}=\frac{10!}{(10-4)!4!}=\frac{10!}{6!4!}=\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}=210\)
Alternatively, the following notation is commonly used in the literature:
\(C_k^n=\begin{pmatrix}
n\\
k
\end{pmatrix}\)
and we read it as "\(n\) choose \(k\)".
The following questions:
• How many ways are there to select 3 people out of 6 for a survey?
• How many ways are there to select 3 pieces of art out 6 to donate?
and many other questions, all have the same answer:
\(C_3^6\)
The following questions:
• How many ways are there to select 5 people out of 9 for a survey?
• How many ways are there to select 5 pieces of art out 9 to donate?
and many other questions, all have the same answer:
\(C_5^9\)
Combinations along with permutations is another building blocks of the combinatorics.