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- https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom)/07%3A_Sets_and_Counting/7.03%3A_PermutationsThere are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the fi...There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. \[7 \mathrm{P} 2=\frac{7 !}{5 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=42 \nonumber \]
- https://math.libretexts.org/Courses/Chabot_College/Math_in_Society_(Zhang)/08%3A_Counting/8.01%3A_The_Fundamental_Counting_PrincipleIf there are \(n_1\) ways to of choosing the first item, \(n_2\) ways of choosing the second item after the first item is chosen, \(n_3\) ways of choosing the third item after the first two have been ...If there are \(n_1\) ways to of choosing the first item, \(n_2\) ways of choosing the second item after the first item is chosen, \(n_3\) ways of choosing the third item after the first two have been chosen, and so on until there are \(n_k\) ways of choosing the last item after the earlier choices, then the total number of choices overall is given by
- https://math.libretexts.org/Courses/Fullerton_College/Math_100%3A_Liberal_Arts_Math_(Claassen_and_Ikeda)/06%3A_Probability/6.04%3A_Counting_MethodsSo far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample...So far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample space we can use counting techniques such as permutations or combinations to count them.
- https://math.libretexts.org/Courses/Las_Positas_College/Math_for_Liberal_Arts/12%3A_Counting/12.01%3A_The_Fundamental_Counting_PrincipleIf there are \(n_1\) ways to of choosing the first item, \(n_2\) ways of choosing the second item after the first item is chosen, \(n_3\) ways of choosing the third item after the first two have been ...If there are \(n_1\) ways to of choosing the first item, \(n_2\) ways of choosing the second item after the first item is chosen, \(n_3\) ways of choosing the third item after the first two have been chosen, and so on until there are \(n_k\) ways of choosing the last item after the earlier choices, then the total number of choices overall is given by
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.03%3A_The_Symmetric_GroupsLet \(\sigma\) and \(\tau\) be defined as follows: \[\sigma= \left ( \begin{array} {ccc} 1&2&3\\ 2&1&3 \end{array} \right ), \quad \quad \tau = \left ( \begin{array} {ccc} 1&2&3\\ 2&3&1 \end{array} \r...Let \(\sigma\) and \(\tau\) be defined as follows: \[\sigma= \left ( \begin{array} {ccc} 1&2&3\\ 2&1&3 \end{array} \right ), \quad \quad \tau = \left ( \begin{array} {ccc} 1&2&3\\ 2&3&1 \end{array} \right )\] It follows that \[\begin{array} {c c c c c c c} \sigma\tau(1) &=&\sigma(\tau(1))&=&\sigma(2)&=1 \\ \sigma\tau(2) &=&\sigma(\tau(2))&=&\sigma(3)&=3\\ \sigma\tau(3) &=&\sigma(\tau(3))&=&\sigma(1)&=2 \end{array}\] Thus we have \[\sigma\tau = \left ( \begin{array} {ccc} 1&2&3\\ 1&3&2 \end{arra…
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Combinatorics_Through_Guided_Discovery_(Bogart)/01%3A_What_is_Combinatorics/1.02%3A_Basic_Counting_PrinciplesIn this section, we explore the basic counting principles through a plethora of examples and exercises. One of our goals in these notes is to show how most counting problems can be recognized as count...In this section, we explore the basic counting principles through a plethora of examples and exercises. One of our goals in these notes is to show how most counting problems can be recognized as counting all or some of the elements of a set of standard mathematical objects. You may have noticed some standard mathematical words and phrases such as set, ordered pair, function, and so on creeping into the problems.
- https://math.libretexts.org/Courses/Chabot_College/Math_in_Society_(Zhang)/08%3A_Counting/8.02%3A_Permutations_and_Combinations\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \tim...\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\[4pt] &=\frac{13 \times 12 \times 11}{3 \times 2 \times 1} \\[4pt] =\frac{1716}{6}=286 \end{align*}\]
- https://math.libretexts.org/Under_Construction/Purgatory/MAT_1320_Finite_Mathematics/05%3A_Sets_and_Counting/5.03%3A_PermutationsThe number of four-letter word sequences is 5P4 = \( \dfrac{5!}{(5-4)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{1} = 5 \cdot 4 \cdot 3 \cdot 2 = 120\) The number of three-letter word sequences is ...The number of four-letter word sequences is 5P4 = \( \dfrac{5!}{(5-4)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{1} = 5 \cdot 4 \cdot 3 \cdot 2 = 120\) The number of three-letter word sequences is 5P3 = \( \dfrac{5!}{(5-3)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1} = 5 \cdot 4 \cdot 3 = 60\) The number of two-letter word sequences is 5P2 = \( \dfrac{5!}{(5-2)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 5 \cdot 4 = 20\)
- https://math.libretexts.org/Courses/University_of_St._Thomas/Math_101%3A_Finite_Mathematics/01%3A_Sets_and_Counting/1.04%3A_Permutations_and_Combinations/1.4.01%3A_PermutationsThere are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the fi...There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. \[7 \mathrm{P} 2=\frac{7 !}{5 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=42 \nonumber \]
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Combinatorics_and_Graph_Theory_(Guichard)/01%3A_Fundamentals/1.08%3A_Sperner's_TheoremWe attempt to partition \(\displaystyle 2^{[n]}\) into \(k={n\choose \lfloor n/2\rfloor}\) chains, that is, to find chains \[\eqalign{ &A_{1,0}\subseteq A_{1,1}\subseteq A_{1,2}\subseteq\cdots\subsete...We attempt to partition \(\displaystyle 2^{[n]}\) into \(k={n\choose \lfloor n/2\rfloor}\) chains, that is, to find chains \[\eqalign{ &A_{1,0}\subseteq A_{1,1}\subseteq A_{1,2}\subseteq\cdots\subseteq A_{1,m_1}\cr &A_{2,0}\subseteq A_{2,1}\subseteq A_{2,2}\subseteq\cdots\subseteq A_{2,m_2}\cr &\vdots\cr &A_{k,0}\subseteq A_{k,1}\subseteq A_{k,2}\subseteq\cdots\subseteq A_{k,m_k}\cr }\nonumber\] so that every subset of \([n]\) appears exactly once as one of the \(\displaystyle A_{i,j}\).
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Discrete_Mathematics_for_Computer_Science_(Fitch)/01%3A_Basics/1.05%3A_Combinatorics-_First_CountsThis page introduces combinatorics, focusing on counting and key terms like "alphabet," "string," and "permutation." It defines "string" as an ordered set of characters and explains "permutation" as a...This page introduces combinatorics, focusing on counting and key terms like "alphabet," "string," and "permutation." It defines "string" as an ordered set of characters and explains "permutation" as any arrangement of those characters. The text provides examples and includes practice checkpoints for generating strings and permutations of varying lengths and characters.
