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Chapter 7: Problem 3

Evaluate the given expression. $$ \frac{5 !}{2 ! 3 !} $$

Short Answer

Expert verified
The expression \(\frac{5!}{2!3!}\) can be solved by first finding the factorials of 5, 2, and 3: 5! = 120, 2! = 2, and 3! = 6. Then, evaluate the expression \(\frac{120}{2 \cdot 6}\), which simplifies to \(\frac{120}{12}\). Finally, simplify the fraction further to get the result: \(\frac{5!}{2!3!} = 10\).

Step by step solution

01

Find the factorials

To find the factorial of a number, you multiply the number by all positive integers less than it. 5! = 5 x 4 x 3 x 2 x 1 = 120 2! = 2 x 1 = 2 3! = 3 x 2 x 1 = 6
02

Evaluate the expression

Now that we have the factorials, we can evaluate the given expression: \(\frac{5!}{2!3!} = \frac{120}{2 \cdot 6} = \frac{120}{12}\)
03

Simplify the fraction

To simplify the fraction, we can divide the numerator and the denominator by their greatest common divisor. In this case, the greatest common divisor of 120 and 12 is 12. Divide both of them by 12: \(\frac{120}{12} = \frac{120\div 12}{12\div 12} = \frac{10}{1}\)
04

Write the final result

Since \(\frac{10}{1}\) is equivalent to 10, our final result is: \(\frac{5!}{2!3!} = 10\)

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Key Concepts

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

Combinatorics
Combinatorics is a fascinating area of mathematics that deals with counting, arrangement, and combination of objects in a set. It is incredibly useful for solving problems related to probability, statistics, and computer science.
Understanding permutations and combinations is key in combinatorics, as these concepts cover different ways of arranging or selecting items.
  • Counting Principle: A fundamental principle in combinatorics, which states that if you have a series of events and each event has a certain number of outcomes, you can multiply the number of outcomes to find the total number of possible outcomes for all events.
  • Factorials: A significant part of combinatorics because they provide a way to calculate the total ways to arrange a set of items. The factorial of a number \( n \) (denoted \( n! \)) is the product of all positive integers up to \( n \).
Combinatorics is not just about theoretical mathematics; it's about application. It helps solve real-world problems by determining the most efficient ways of arranging, organizing, and managing data.
Permutation and Combination
Permutation and combination are two fundamental concepts in the study of combinatorics. They deal with the different ways to select items from a collection.
  • Permutation: This refers to the arrangement of items in a specific order. If the order matters, it's a permutation. For example, the sequence ABC is different from BAC in permutations.
  • Combination: This is about selecting items where the order does not matter. For example, choosing two fruits from a set of apples, oranges, and bananas. Whether you pick an apple and then an orange, or an orange and then an apple, it counts as the same combination.
Let's relate these concepts to factorials, which were used in our original exercise. When calculating combinations, we use the formula:\[ C(n, r) = \frac{n!}{r!(n-r)!} \]Here, \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. Factorials thus play a crucial part in determining the number of combinations, simplifying the process by which they are calculated.
Mathematical Expressions Simplification
Simplifying mathematical expressions is a fundamental skill in algebra and beyond. It involves rewriting expressions in a simpler form while retaining their original values.
  • In the original exercise, simplifying the fraction \( \frac{120}{12} \) by dividing both numerator and denominator by their greatest common divisor is an example of simplification.
  • Simplification often includes reducing fractions, combining like terms, or factoring where applicable.
  • The aim is to make complex expressions easier to work with, especially in calculations or further algebraic manipulations.
For instance, in the expression from the exercise, simplifying \( \frac{5!}{2!3!} \) involved calculating factorials first, then dividing by the greatest common divisor.
Understanding how to simplify expressions allows easier navigation through mathematical problems, enhancing both computational and problem-solving efficiency.

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Most popular questions from this chapter

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The following table, compiled in 2004 , gives the percentage of music downloaded from the United States and other countries by U.S. users: $$ \begin{array}{lcccccccc} \hline \text { Country } & \text { U.S. } & \text { Germany } & \text { Canada } & \text { Italy } & \text { U.K. } & \text { France } & \text { Japan } & \text { Other } \\ \hline \text { Percent } & 45.1 & 16.5 & 6.9 & 6.1 & 4.2 & 3.8 & 2.5 & 14.9 \\\ \hline \end{array} $$ a. Verify that the table does give a probability distribution for the experiment. b. What is the probability that a user who downloads music, selected at random, obtained it from either the United States or Canada? c. What is the probability that a U.S. user who downloads music, selected at random, does not obtain it from Italy, the United Kingdom (U.K.), or France?
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