91Ó°ÊÓ

Stirling numbers of the second kind, denoted as \(S(n, m)\), are essential in combinatorics. They represent the number of ways to partition a set of \(n\) elements into \(m\) non-empty subsets. For example, if you have a set of 7 elements and you want to know how many ways you can divide this set into 5 non-empty subsets, you'd use \(S(7, 5)\). This helps because it gives you a structured way to break down the problem of creating specific subsets.
In the context of our problem, we used \(S(7, 5) = 1407\). This provides the count of ways we can split 7 elements into 5 groups, ensuring each group has at least one element. This is a fundamental part of calculating onto functions, as we'll see when we consider permutations with factorials.

The factorial of a non-negative integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). Factorials are vital in permutations and combinations, often simplifying complex counting problems.
For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). This describes the number of ways to arrange 5 elements. In our problem, after determining the number of partitions using Stirling numbers, we need the factorial of 5 (\(5!\)) to apply permutations to each partition. This step is crucial because it considers every possible mapping to form onto functions.
By multiplying \(S(7, 5) = 1407\) with the factorial of 5, we correctly account for all permutations in our problem's solution.

A surjective function (or onto function) ensures that every element in the target set is mapped to by at least one element in the source set. For our specific problem, it means every element in set B (with 5 elements) must be the image of at least one element from set A (with 7 elements).
To count the number of surjective functions, we use the formula \(S(n, m) \times m!\). Here, \(S(n, m)\) accounts for partitioning, ensuring each group is represented, while \(m!\) handles permutations.
By reviewing our solution, where we calculated \(S(7, 5) = 1407\) and multiplied it by \(5! = 120\), yielding \(168,840\), we ensured that every target element had at least one pre-image. This guarantees a complete surjection, fulfilling the requirements for an onto function.