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Consider functions \(f:\\{1,2,3,4\\} \rightarrow\\{1,2,3,4,5,6\\}\) (a) How many functions are there total? (b) How many functions are injective? (c) How many of the injective functions are increasing? To be increasing means that if \(a

Short Answer

Expert verified
There are a total of 1296 functions, 360 of which are injective, and only 15 of the injective functions are increasing.

Step by step solution

01

- Total Number of Functions

Since a function from a set with 4 elements to a set with 6 elements can assign each element in the domain to any one of the 6 elements in the codomain, for each of the 4 elements in the domain we have 6 choices. The total number of functions is found by multiplying the number of choices for each element, which is 6^4.
02

- Number of Injective Functions

An injective function, also known as a one-to-one function, assigns distinct elements in the domain to distinct elements in the codomain. Since the domain has 4 elements and the codomain has 6 elements, we can choose 4 distinct elements for the images in 6P4 ways (permutations of 6 elements taken 4 at a time), which is calculated as 6! / (6-4)!.
03

- Number of Increasing Injective Functions

An increasing injective function assigns larger outputs for larger inputs without repetition. Since we can't assign the same output to two different inputs and the function must be increasing, for the range we are just choosing 4 increasing elements out of 6, which is the number of combinations of 6 elements taken 4 at a time, denoted as 6C4, calculated as 6! / (4! * (6-4)!).

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Key Concepts

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

Injective Function
When studying discrete mathematics, a crucial concept to understand is the injective function, also known as a one-to-one function. Imagine each input in a function as a unique key to a lock—every key fits only one specific lock and no two keys can open the same lock. Similarly, in an injective function, each element of the domain ('input' set) is paired with a unique element in the codomain ('output' set), ensuring that no two different inputs share the same output.

In our exercise, when forming injective functions from a set of 4 elements \( \{1,2,3,4\} \) to a set of 6 elements \( \{1,2,3,4,5,6\} \), this uniqueness means we must select 4 distinct outputs for our inputs from the larger set. This is done through permutations. Think of permutations as ways to arrange books on a shelf—how many different orders can you create with four unique books out of six? The math boils down to 6P4, which is shorthand for '6 permute 4'. This is calculated using the factorial function \( 6! / (6-4)! \) and gives us the number of all possible one-to-one mappings.

It's essential to understand that injective functions are all about distinct pairings. If a student is asked to create or identify an injective function, they must ensure each input has its own exclusive output. This is a core principle in not only discrete mathematics but also in fields like computer science where unique identifiers are key.
Increasing Function
An increasing function in discrete mathematics is much like climbing a staircase—each step takes you higher than the last. More formally, it means that for any two elements \( a \) and \( b \) in the domain, if \( a < b \) then \( f(a) < f(b) \). In the context of the exercise, where we have the function \( f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\} \), an increasing function will always assign a larger output to a larger input without exception. This orderly progression gives structure to the function and has profound implications in sequencing and sorting algorithms.

Visualizing this, you can imagine each choice of output as a higher step that must follow the preceding one. No steps can be on the same level (that would violate injectivity) or decrease (that would violate the increasing property). Given our set, we want to find the number of ways we can choose 4 increasing outputs from 6 options. Using combinations, denoted as 6C4, we find the number of ways to arrange 4 elements in increasing order from a set of 6. This is computed via \( 6! / (4! * (6-4)!) \).

Understanding increasing functions are essential in mathematical sequences, computer algorithms, and any context where order and growth are integral. This concept ensures that each input not only maps to a unique output but does so in a manner that maintains a consistent upward trend.
Function Permutations
Permutations play a pivotal role in discrete mathematics, especially when it comes to understanding functions. A permutation is essentially an ordered combination, like a specific sequence of numbers drawn in a lottery. In the realm of functions, when we talk about function permutations, we're interested in the different ways we can arrange the mapping of inputs to outputs where order matters.

In the provided exercise, for injective functions between the sets \( \{1,2,3,4\} \) and \( \{1,2,3,4,5,6\} \), we use the concept of permutations to determine the number of distinct ways we can pair these elements. Using the permutation formula, 6P4, we calculate the total number of permutations by using factorials. The factorial function symbol \( ! \) denotes the product of all positive integers up to that number. So, \( 6! \) means \( 6 \times 5 \times 4 \times 3 \times 2 \times 1 \). The permutation formula, \( \frac{6!}{(6-4)!} \) gives us the total different arrangements of selecting 4 items out of 6, each arrangement having a unique order.

This calculation is paramount for understanding not only functions but any scenario in mathematics or computer science that requires the enumeration of different sequential arrangements. By mastering the concept of function permutations, students gain insight into more complex topics such as combinations, probability, and statistical analysis where the order of elements is a determining factor.

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Most popular questions from this chapter

A combination lock consists of a dial with 40 numbers on it. To open the lock, you turn the dial to the right until you reach a first number, then to the left until you get to second number, then to the right again to the third number. The numbers must be distinct. How many different combinations are possible?

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