Question

This question concerns functions \(f:\\{A, B, C, D, E\\} \rightarrow\\{1,2,3,4,5,6,7\\} .\) How many such functions are there? How many of these functions are injective? How many are surjective? How many are bijective?

Short answer

There are \(7^5\) total functions, \(7*6*5*4*3\) injective functions, \(7^5 - \binom{7}{1}*6^5 + \binom{7}{2}*5^5 - \binom{7}{3}*4^5 + \binom{7}{4}*3^5 - \binom{7}{5}*2^5\) surjective functions, and 0 bijective functions.

Step-by-step solution

Count Total Functions

Total functions can be counted by assigning to each of the 5 elements in the domain one of the 7 elements in the codomain. This gives a total of \(7^5\) functions.

Count Injective Functions

Injective functions can be counted by assigning to each of the 5 elements in the domain one of the 7 elements in the codomain such that no two domain elements map to the same codomain element. This is computed as the permutation \(P(7,5)\) which equals \(7*6*5*4*3\).

Count Surjective Functions

Surjective functions are more complex to count. First, some sets will not have any surjections if the codomain has more elements than the domain. To calculate the number of surjections, we will use the principle of inclusion-exclusion. This gives us \(7^5 - \binom{7}{1}*6^5 + \binom{7}{2}*5^5 - \binom{7}{3}*4^5 + \binom{7}{4}*3^5 - \binom{7}{5}*2^5\).

Count Bijective Functions

A function is bijective if and only if it is both injective and surjective. However, in this case, since the domain has less elements than the codomain, it is impossible to construct a bijective function from A to B, because there will always be elements in the codomain which will not get mapped by any element from the domain.

Key concepts

Injective Functions

Imagine you attend a unique potluck dinner where each guest brings a different dish, and no two guests bring the same one. This situation is similar to injective functions, also known as one-to-one functions. In mathematics, an injective function ensures that each element in the domain (the potluck guests) corresponds to a unique element in the codomain (the variety of dishes).

Let's put this into a formulaic perspective. A function f from A to B being injective means formally that if f(a) = f(b) then a must equal b. Another way to think of it is that every piece of input into the function has a distinct output, just like at our potluck, every guest's dish can be uniquely identified.

Using the step-by-step solution provided, counting injective functions from a set with 5 elements to a set with 7 elements involves assigning a unique codomain element to each domain element. Hence, you pick different elements without replacement. For the first domain element, you have 7 choices, for the second, 6, and it goes down until the fifth has only 3 choices. The product of these choices gives you the total number of possible injective functions, computed as a permutation P(7,5).

Surjective Functions

Now, imagine a buffet where every dish must be tried at least once. No dish must feel left out! This is the essence of surjective functions, or onto functions. A surjective function from domain A to codomain B guarantees that every element in the codomain has at least one corresponding element in the domain. In other words, every dish at the buffet will satisfy at least one guest's palette.

In terms of equations, for every element y in the codomain B, there is at least one element x in the domain A such that f(x) = y. To count the number of surjective functions in the exercise, one can't just allocate the elements directly because overcounts and undercounts need to be considered. Instead, we use a mathematical approach called the principle of inclusion-exclusion. It allows us to consider all possible distributions of elements in the codomain while subtracting the arrangements that leave out one or more elements—ensuring every element from the buffet is indeed tasted.

However, it's noteworthy to mention that surjective functions between sets where the domain has fewer elements than the codomain do not exist, as there can't be enough elements to map to each element of the codomain at least once.

Bijective Functions

Combining the ideas of both injective and surjective functions gives us bijective functions, the superstars of functions. In a bijective function, every domain element pairs up with a unique codomain element, and each codomain element is used exactly once, similar to a perfect dance where every dancer has a partner, and no one is left out.

In formal terms, a bijective function is one where for every element a in the domain A, there is a unique element b in the codomain B and vice versa, so that f(a) = b. The exercise solution tells us that a bijection is not possible if there are more elements in the codomain than in the domain; it's like having more guests than seats at a dinner - someone will always be standing.

This means there are no bijective functions in the given problem, as the condition for bijectivity is not met. The special case of such a perfect matching between the sets requires them to have an equal number of elements, ensuring all criteria of both injectivity and surjectivity are satisfied.