Section 5.3.H: Homework
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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 5.3: Permutations - Homework Exercises
You have 5 different books and want to arrange them on a shelf. How many different arrangements are possible?
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5! = 120 different arrangements
A baseball team has 9 players, and each player must be assigned a position in the batting lineup. How many different batting orders can be created using the 9 players?
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9! = 362,880 different batting orders
Six runners compete in a race. Assume there are no ties. How many different possible finish orders are there?
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6! = 720 different finishes
A locker requires a 3-number combination where each number is from 1 to 40 and no number can be repeated. How many different combinations are possible?
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\(_{40}P_3=\) 59,280 different ways
A class of 25 students needs to elect a President, Vice President, Secretary, and Treasurer. No student can hold more than one position. How many different ways can these four positions be filled?
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\(_{25}P_4=\) 303,600 different ways
A baseball team has 15 players, but only 9 can be in the starting lineup. The order in the lineup matters. How many different batting orders are possible?
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\(_{15}P_9=\) 1,816,214,400 different orders
Eight runners compete in a race. How many different ways can they finish 1st, 2nd, and 3rd place? (Assume no ties)
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\(_8P_3=\) 336 different finishes
How many 5-letter "words" (they don't need to be real words) can be formed from the 26 letters of the alphabet, if no letter can be used more than once?
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\(_{26}P_5=\) 7,893,600 different words
How many distinct arrangements can be made with the letters in the word "MATH"?
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4! = 24 different arrangements
How many distinct arrangements can be made with the letters in the word "UNCOPYRIGHTABLE"?
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15! = 1,307,674,368,000 different arrangements
How many distinct arrangements can be made with the letters in the word "GREEN"?
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\(\frac{5!}{2!}=\frac{120}{2}=\) 60 different arrangements
How many distinct arrangements can be made with the letters in the word "BANANA"?
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\(\frac{6!}{3!\cdot2!}=\frac{720}{6\cdot2}=\frac{720}{12}=\) 60 different arrangements
How many distinct arrangements can be made with the letters in the word "BOOKKEEPER"?
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\(\frac{10!}{3!\cdot2!\cdot2!}=\frac{3,628,800}{6\cdot2\cdot2}=\frac{3,628,800}{24}=\) 151,200 different arrangements
How many distinct arrangements can be made with the letters in the word "POSSESSES"?
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\(\frac{9!}{5!\cdot2!}=\frac{362,880}{120\cdot2}=\frac{362,880}{240}=\) 1,512 different arrangements
The word "SUPERCALIFRAGILISTICEXPIALIDOCIOUS" was popularized by the 1964 Disney film Mary Poppins, where it appears in a famous song of the same name sung by Julie Andrews (Mary Poppins) and Dick Van Dyke (Bert). How many distinct arrangements can be made with the letters in the word "SUPERCALIFRAGILISTICEXPIALIDOCIOUS"?
- Answer
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\(\frac{34!}{7!\cdot3!\cdot3!\cdot2!\cdot2!\cdot2!\cdot2!\cdot2!}≈ 7.63 × 10^{36}\) different arrangements