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Chapter 6: Problem 33

$$ \text { Write all the } 3 \text {-permutations of }\\{s, t, u, v\\} \text {. } $$

Short Answer

Expert verified
The 24 3-permutations of the set {s, t, u, v} are: stu, stv, suv, svt, svt, sut, tsu, tsv, tus, tvs, tvu, tuv, ust, usv, uts, utv, uvt, uvs, vst, vsu, vts, vtu, vus, and vut.

Step by step solution

01

Understanding permutation

A permutation is an arrangement of objects in a specific order. In this case, we want to find the arrangements of three elements from the given set {s, t, u, v}. There are 4 elements in the set, and we want to arrange them 3 at a time. This can be represented as P(4, 3).
02

Calculate the number of 3-permutations

To calculate the number of 3-permutations, we can use the formula: \(P(n, r) = \frac{n!}{(n-r)!}\) Here, n = 4 (total number of elements) and r = 3 (number of elements to arrange). Plugging in the values, we get: \(P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = \frac{4 脳 3 脳 2 脳 1}{1} = 24\) So, there are 24 possible 3-permutations of the set {s, t, u, v}.
03

List all the 3-permutations

Now, let's list all the 3-permutations. We can do this by fixing one element at a time and listing the permutations of the other two elements. 1. Fix s first: stu, stv, suv, svt, svt, sut 2. Fix t first: tsu, tsv, tus, tvs, tvu, tuv 3. Fix u first: ust, usv, uts, utv, uvt, uvs 4. Fix v first: vst, vsu, vts, vtu, vus, vut In total, we have found 24 3-permutations, which matches the result calculated in step 2.

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Key Concepts

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

Arrangement of Objects
Understanding how to arrange objects is a fundamental concept in permutations. In a permutation, the order of the objects matters. For instance, arranging the letters {s, t, u} is different from arranging them as {t, s, u}. When we talk about arrangements, we're organizing these objects in a sequence where each sequence is unique due to the order of elements.
Consider the task of arranging three letters from a set of four: {s, t, u, v}. Here, we choose any three letters and organize them in every possible order. This way, the specific order creates different arrangements.
Permutation Formula
The permutation formula plays a crucial role in calculating how we can arrange objects. It is represented as:
\(P(n, r) = \frac{n!}{(n-r)!} \)
In this formula, \(n\) is the total number of objects available, and \(r\) is the number of objects we want to arrange. The symbol \(!\) denotes a factorial, meaning a series of multiplying descending numbers.
  • For \(n = 4\) and \(r = 3\), our calculation becomes \(P(4, 3) = \frac{4!}{1!} = 24\).
  • Here \(4! = 4 \times 3 \times 2 \times 1\), representing all ways to arrange four items, and \(1!\) is just 1.
This result confirms there are 24 ways to arrange three letters chosen from the set {s, t, u, v}.
Combinatorics
Combinatorics deals with counting, arranging, and understanding the deeper properties of sets of objects. Permutations are a part of combinatorics.
When dealing with problems like arranging 3 elements from a set of 4, combinatorics helps us understand not just the quantity but also the logical structure of these arrangements.
  • It organizes problems into solvable steps using tools like the permutation formula.
  • By understanding combinatorics, solving arrangement problems becomes more systematic as it provides a framework to count and arrange items without errors.
With this knowledge, breaking down complex arrangement tasks into smaller, manageable parts becomes feasible, enabling efficient solutions.

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Most popular questions from this chapter

Express each of the sums in \(24-35\) in closed form (without using a summation symbol and without using an ellipsis \(\cdots\) ). $$ \sum_{i=0}^{m}\left(\begin{array}{c} m \\ i \end{array}\right) p^{m-i} q^{2 i} $$

A pair of fair dice, one blue and the other gray, are rolled. Let \(A\) be the event that the number face up on the blue die is 2 , and let \(B\) be the event that the number face up on the gray die is 4 or 5 . Show that \(P(A \mid B)=P(A)\) and \(P(B \mid A)=P(B)\).

Prove Bayes' Theorem for \(n=2\). That is, prove that if a sample space \(S\) is a union of mutually disjoint events \(B_{1}\) and \(B_{2}\), if \(A\) is an event in \(S\) with \(P(A) \neq 0\), and if \(k=1\) or \(k=2\), then $$ P\left(B_{k} \mid A\right)=\frac{P\left(A \mid B_{k}\right) \cdot P\left(B_{k}\right)}{P\left(A \mid B_{1}\right) \cdot P\left(B_{1}\right)+P\left(A \mid B_{2}\right) \cdot P\left(B_{2}\right)} $$

An alternative way to derive Theorem 6.4.1 uses the following division rule: Let \(n\) and \(k\) be integers so that \(k\) divides \(n\). If a set consisting of \(n\) elements is divided into subsets that each contain \(k\) elements, then the number of such subsets is \(n / k\). Explain how Theorem \(6.4 .1\) can be derived using the division rule.

a. How many integers are there from 1000 through 9999 ? b. How many odd integers are there from 1000 through \(9999 ?\) c. How many integers from 1000 through 9999 have distinct digits? d. How many odd integers from 1000 through 9999 have distinct digits? e. What is the probability that a randomly chosen four-digit integer has distinct digits? has distinct digits and is odd?
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