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Wie viele verschiedene - nicht notwendig sinnvolle - Worte kann man aus allen Buchstaben der folgenden Worte bilden? (a) dort, (b) gelesen, (c) Ruderregatta.

Short Answer

Expert verified
(a) dort (b) gelesen (c) Ruderregatta Answer: (a) dort: 24, (b) gelesen: 2520, and (c) Ruderregatta: 19,958,400 different words can be formed.

Step by step solution

01

Count letters and find repetitions for each word

First, we find out the total number of letters and the multiplicities of each letter for each given word. (a) dort: Total letters = 4 (d, o, r, t) No letter is repeated. (b) gelesen: Total letters = 7 (g, e, l, s, e, n) Repetition: e repeated 2 times. (c) Ruderregatta: Total letters = 12 (R, u, d, e, r, r, e, g, a, t, t, a) Repetition: R repeated 1 time, u repeated 1 time, d repeated 1 time, e repeated 2 times, r repeated 3 times, g repeated 1 time, a repeated 2 times, t repeated 2 times.
02

Calculate permutations for each word

Now, we will use the formula for permutations with repetition to determine the number of different words that can be formed for each case. (a) dort: Number of permutations = 4! / (1! * 1! * 1! * 1!) = 24 / 1 = 24. (b) gelesen: Number of permutations = 7! / (1! * 2! * 1! * 1! * 1!) = 5040 / 2 = 2520. (c) Ruderregatta: Number of permutations = 12! / (1! * 1! * 1! * 2! * 3! * 1! * 2! * 2!) = 479001600 / (2 * 6 * 2 * 2) = 19958400. So, the numbers of different words that can be formed from the given words are: (a) dort: 24 (b) gelesen: 2520 (c) Ruderregatta: 19958400

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Key Concepts

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

Permutation Formula
The concept of permutations is instrumental in the study of probability and combinatorics, where it is important to count distinct arrangements of set elements. The permutation formula is a way to find the number of possible arrangements (permutations) of a certain number of objects within a set, where the order of these objects matters.

When all the objects under consideration are distinct, the number of permutations of n objects is given by the factorial of n, denoted as n!. However, when there are repetitions, as is common in real-world examples like the arrangement of letters in words, the formula is adjusted to divide by the factorials of the frequencies of the repeating elements. In mathematical terms, if n is the total number of objects and there are n1, n2, ..., nk repetitions, the permutation formula with repetition is \[ \frac{n!}{n1! \times n2! \times ... \times nk!} \].

This allows for a fair count by considering arrangements that look different but are essentially the same due to the repetition of identical elements to be counted as a single unique arrangement.
Factorial
The factorial of a non-negative integer n, denoted as n!, is the product of all positive integers less than or equal to n. For instance, 5! equals to 5 × 4 × 3 × 2 × 1 which is 120. The factorial operation is fundamental in combinatorics, particularly in counting permutations and combinations where order and selection are key factors.

An important point to note is that the factorial of zero, 0!, is always defined to be 1, which often comes as a surprise to those new to the concept. This may seem counterintuitive at first but is necessary for the math of combinatorics to work correctly, particularly in cases involving zero objects where there's essentially one way to arrange nothing.
Combinatorics
Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is related to all other fields of mathematics, and has applications in computer science, physics, and statistics, among other areas.

In combinatorics, we often deal with permutations and combinations to find the number of ways objects can be arranged or selected. Permutations with repetition, as shown in the exercise, are a critical subset of this field, allowing for a more nuanced count of arrangements when identical items are present. By building a strong foundation in understanding permutations and their calculations, students gain the tools necessary to tackle complex problems in probability, game theory, and beyond.

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Most popular questions from this chapter

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