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Exercises \(67-72\) are based on the following jokes about books: \(\cdot\) "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." - Groucho Marx \(\cdot\) "I recently bought a book of free verse. For \(\$ 12\)." \- George Carlin \(\cdot\) "If a word in the dictionary was misspelled, how would we know?" - Steven Wright \(\cdot\) "Encyclopedia is a Latin term. It means 'to paraphrase a term paper." - Greg Ray \(\cdot\) "A bookstore is one of the only pieces of evidence we have that people are still thinking." - Jerry Seinfeld \(\cdot\) "I honestly believe there is absolutely nothing like going to bed with a good book. Or a friend who's read one." \(-\)Phyllis Diller In how many ways can these six jokes be ranked from best to worst?

Short Answer

Expert verified
There are 720 ways to rank the six jokes from best to worst.

Step by step solution

01

Understand Permutations

In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. Because each joke can only be used once and the order in which the items are arranged matters, this problem is a permutation.
02

Use Permutation Formula

The formula for permutations is \(nPr = n!/(n-r)!\), where \(n\) is the number of items to choose from, \(r\) is how many items are chosen, and \( ! \) denotes factorial, which is the product of an integer and all the integers below it (e.g., \(5! = 5 × 4 × 3 × 2 × 1 = 120\)). In this case, \(n\) and \(r\) are both 6, because all six jokes are being ranked.
03

Calculate the Permutation

To calculate the permutation, we substitute \(n = 6\) and \(r = 6\) into the permutation formula: \(nPr = 6!/(6-6)!\). Simplifying that gives \(6!/0!\), and because \(0! = 1\), the result is \(6 != 720\). These calculations indicate that there are 720 ways to rank the six jokes from best to worst.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorials
A factorial is a mathematical operation that is fundamental in understanding permutations and combinations. It is typically denoted by an exclamation mark '!' after a number. For any given positive integer, say 6, the factorial of this number, written as 6!, represents the product of all positive integers less than or equal to 6. Thus, we have:\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]Factorials grow very quickly with larger numbers, as each new integer is multiplied by all the integers below it. They are crucial in calculating permutations since permutations require multiplying sets of descending integers to determine the number of possible arrangements.Some key points to note about factorials include:
  • 0! is defined as 1. This is a fundamental property used in permutations and combinations.
  • The factorial function is only defined for non-negative integers.
Permutation Formula
The permutation formula is essential for calculating different ways to order a set of items where the sequence matters. The formula is given as:\[ nPr = \frac{n!}{(n-r)!} \]Here, \( n \) represents the total number of items from which you choose, \( r \) symbolizes the number of items you want to arrange, and the '!' symbolizes factorial. The formula accounts for every possible order by considering all items, then removing combinations where the chosen items \( r \) overlap.In the context of ranking six jokes, both \( n \) and \( r \) are 6, since we are interested in arranging all the items. Applying this to our case:\[ 6P6 = \frac{6!}{(6-6)!} = \frac{6!}{0!} = 720 \]Given that 0! equals 1, simplifying gives the number of permutations as 720. Thus, there are 720 different ways to rank the six jokes from best to worst.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of objects. It provides various tools and methods to calculate probabilities and possibilities in complex scenarios, such as ranking or arranging items. In combinatorics, you'll encounter key concepts like:
  • Permutations, where the arrangement order matters, as we see with the joke ranking problem.
  • Combinations, where the order does not matter, typically used when grouping items without concern for sequence.
While combinatorics simplifies understanding and solving problems involving arrangements and selections, it can initially seem overwhelming due to numerous formulas and principles. However, with practice and an understanding of factorials and permutation formulas, it becomes more approachable and immensely useful for problem-solving. Learning combinatorics involves recognizing the conditions of your problem. Is the order important? If so, permutations are your tool. If not, combinations provide the solution aid, allowing both students and professionals to make accurate calculations efficiently in diverse scenarios.

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Most popular questions from this chapter

Exercises \(67-72\) are based on the following jokes about books: \(\cdot\) "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." - Groucho Marx \(\cdot\) "I recently bought a book of free verse. For \(\$ 12\)." \- George Carlin \(\cdot\) "If a word in the dictionary was misspelled, how would we know?" - Steven Wright \(\cdot\) "Encyclopedia is a Latin term. It means 'to paraphrase a term paper." - Greg Ray \(\cdot\) "A bookstore is one of the only pieces of evidence we have that people are still thinking." - Jerry Seinfeld \(\cdot\) "I honestly believe there is absolutely nothing like going to bed with a good book. Or a friend who's read one." \(-\)Phyllis Diller If the order in which these jokes are told makes a difference in terms of how they are received, how many ways can they be delivered if a joke by a man is told first?

Follow the outline below and use mathematical induction to prove the Binomial Theorem: $$\begin{aligned}(a+b)^{n} &-\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\&+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}$$ a. Verify the formula for \(n-1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$\begin{array}{l}(a+b)^{k+1}-\left(\begin{array}{c}k \\\0\end{array}\right)a^{k+1}+\left[\left(\begin{array}{c}k \\\0\end{array}\right)+\left(\begin{array}{c}k \\\1\end{array}\right)\right] a^{k} b \\\\+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}\end{array}$$ e. Use the result of Exercise 84 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) $$\begin{aligned}&-\left(\begin{array}{l}n+1 \\\r+1\end{array}\right), \text { then }\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)-\left(\begin{array}{c}k+1 \\\1\end{array}\right) \text { and }\\\&\left(\begin{array}{l}k \\\1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)-\left(\begin{array}{c}k+1 \\\2\end{array}\right)\end{aligned}$$ f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)-\left(\begin{array}{c}k+1 \\ 0\end{array}\right)(\text { why? })\) and \(\left(\begin{array}{l}k \\\ k\end{array}\right)-\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

Five men and five women line up at a checkout counter in a store. In how many ways can they line up if the first person in line is a woman and the people in line alternate woman, man, woman, man, and so on?

Exercises will help you prepare for the material covered in the next section. $$\text { Simplify: } \frac{k(k+1)(2 k+1)}{6}+(k+1)^{2}.$$

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