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Determine the number of 11 -permutations of the multiset $$ S=\\{3 \cdot a, 4 \cdot b, 5 \cdot c\\} . $$

Short Answer

Expert verified
The total number of 11-permutations of the multiset \(S = \{3 \cdot a, 4 \cdot b, 5 \cdot c\}\) is given by the sum \(\frac{11!}{2!4!5!} + \frac{11!}{3!3!5!} + \frac{11!}{3!4!4!} \).

Step by step solution

01

Identify Possible Combinations

First, we recognize that since we are looking for 11-character permutations out of a 12-character set, there are three possibilities: excluding one 'a', one 'b', or one 'c'. Thus, we will have either: MultiSet 1 with 2 'a', 4 'b', 5 'c', MultiSet 2 with 3 'a'-, 3 'b', 5 'c' or MultiSet 3 with 3 'a', 4 'b', and 4 'c'.
02

Calculate Permutations of Each Set

Next, we calculate the permutations of each of these new multisets. For a multiset with m elements (with \(n_1, n_2, …, n_k\) copies of each of its distinct elements), the number of r-permutations is given by the multinomial coefficient: _mPr_ = \(\frac{m!}{n_1!n_2!...n_k!}\). Using this formula, we find the number of 11-permutations of each multiset: MultiSet 1: \(\frac{11!}{2!4!5!} \), MultiSet 2: \(\frac{11!}{3!3!5!} \), MultiSet 3: \(\frac{11!}{3!4!4!} \).
03

Sum The Permutation Numbers

Finally, we add these three numbers together to obtain the total number of 11-permutations of the original multiset S. The total number of permutations is therefore: \(\frac{11!}{2!4!5!} + \frac{11!}{3!3!5!} + \frac{11!}{3!4!4!}\).

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Key Concepts

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

Combinatorics
At its core, combinatorics is a branch of mathematics focused on counting, arranging, and combining objects in a structured way. It plays a crucial role in several areas of mathematics and has applications in computer science, physics, and statistics. Combinatorics helps us understand the possibilities of different configurations that can be formed from a given set of objects.

For instance, if we wish to determine how many different ways we can arrange books on a shelf or select a committee from a group of people, combinatorics provides the methods to calculate these scenarios. It covers various principles and techniques, such as the rule of sum, the rule of product, permutations and combinations, which are foundational for solving complex problems involving multiset permutations as seen in the given exercise.
Permutations and Combinations
The concepts of permutations and combinations are central to the study of combinatorics.

Permutations focus on the arrangement of a set of objects in a specific order. For example, if you want to know in how many different ways you can arrange the letters A, B, and C, you are calculating permutations. If the order matters, it is a permutation.

Combinations, on the other hand, are about selecting items from a group where order does not matter. If you were picking 2 fruits out of an assortment of 5, and you didn't care about the order in which you chose them, you'd be calculating combinations. It is essential to understand the difference between these two concepts to apply the correct mathematical formulas and techniques to counting problems.
Multinomial Coefficients
In the world of combinatorics, multinomial coefficients expand upon the idea of combinations to multiset permutations where repetition of elements is allowed. The multinomial coefficient is denoted by the formula: \[\binom{m}{n_1,n_2,...,n_k} = \frac{m!}{n_1!n_2!...n_k!}\] where \(m!\) is the factorial of the total number of items, and \(n_1!, n_2!, ..., n_k!\) are the factorials of the counts of each distinct item.

This coefficient determines the number of ways to divide a set into smaller subsets of specified sizes. In essence, it generalizes the concept of combinations by allowing more than just two subsets. This is especially useful when dealing with multisets like the one in the example provided, where we have a collection of objects with duplicates.
Factorial Notation
The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). For instance, \(5!\) can be calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorial notation is a mathematical shorthand used extensively in combinatorics to simplify expressions of permutations and combinations.

The factorial function grows very fast, which means even for small values of \(n\), the value of \(n!\) can be quite large. It is important to understand how to manipulate factorials when evaluating multinomial coefficients or solving permutation and combination problems, as incorrect use can lead to vastly incorrect answers. In problems like the exercise provided, factorials allow us to calculate the total number of arrangements systematically and efficiently.

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Most popular questions from this chapter

A committee of five people is to be chosen from a club that boasts a membership of \(10 \mathrm{men}\) and 12 women. How many ways can the committee be formed if it is to contain at least two women? How many ways if, in addition, one particular man and one particular woman who are members of the club refuse to serve together on the committee?

A bakery sells six different kinds of pastry. If the bakery has at least a dozen of each kind, how many different options for a dozen of pastries are there? What if a box is to contain at least one of each kind of pastry?

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A football team of 11 players is to be selected from a set of 15 players, 5 of whom can play only in the backfield, 8 of whom can play only on the line, and 2 of whom can play either in the backfield or on the line. Assuming a football team has 7 men on the line and 4 men in the backfield, determine the number of football teams possible.

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