Solving the Job Assignment Problem with Constraints

The question asks for the number of ways to assign 5 different jobs to 4 different employees such that each employee receives at least one job. This is a classic combinatorics problem involving surjective mappings or the Principle of Inclusion-Exclusion.

Understanding the Problem Setup

• We have \(m=5\) distinct jobs.
• We have \(n=4\) distinct employees.
• Each job must be assigned to exactly one employee.
• Every employee must be assigned at least one job.

This problem is equivalent to finding the number of surjective functions from a set of 5 elements (jobs) to a set of 4 elements (employees).

Method 1: Using the Principle of Inclusion-Exclusion (PIE)

First, let's find the total number of ways to assign 5 different jobs to 4 different employees without any restrictions. Each of the 5 jobs can be assigned to any of the 4 employees independently. So, the total number of unrestricted assignments is:

Now, we need to subtract the cases where at least one employee is not assigned any job. We use the Principle of Inclusion-Exclusion for this. Let \(E_i\) be the property that employee \(i\) is not assigned any job.

The number of ways where at least one employee is not assigned is given by:

Let's calculate each term:

• \( \sum |E_i| \): Choose 1 employee not to be assigned (\(\binom{4}{1}\) ways). The 5 jobs must be assigned to the remaining 3 employees. There are \(3^5\) ways for this. \[ \binom{4}{1} \times 3^5 = 4 \times 243 = 972 \]
• \( \sum |E_i \cap E_j| \): Choose 2 employees not to be assigned (\(\binom{4}{2}\) ways). The 5 jobs must be assigned to the remaining 2 employees. There are \(2^5\) ways for this. \[ \binom{4}{2} \times 2^5 = 6 \times 32 = 192 \]
• \( \sum |E_i \cap E_j \cap E_k| \): Choose 3 employees not to be assigned (\(\binom{4}{3}\) ways). The 5 jobs must be assigned to the remaining 1 employee. There is \(1^5\) way for this. \[ \binom{4}{3} \times 1^5 = 4 \times 1 = 4 \]
• \( |E_1 \cap E_2 \cap E_3 \cap E_4| \): Choose 4 employees not to be assigned (\(\binom{4}{4}\) ways). The 5 jobs must be assigned to the remaining 0 employees. There are \(0^5 = 0\) ways for this. \[ \binom{4}{4} \times 0^5 = 1 \times 0 = 0 \]

Using PIE, the number of assignments where at least one employee is not assigned is:

The number of ways to assign the 5 jobs to 4 employees such that every employee is assigned at least 1 job is the total number of assignments minus the number of assignments where at least one employee is not assigned:

Method 2: Using Stirling Numbers of the Second Kind

The number of surjective functions from a set of size \(m\) to a set of size \(n\) is given by \(n! S(m, n)\), where \(S(m, n)\) is the Stirling number of the second kind. \(S(m, n)\) counts the number of ways to partition a set of \(m\) elements into \(n\) non-empty subsets.

In this problem, we are mapping 5 jobs (elements) to 4 employees (subsets). We first partition the 5 distinct jobs into 4 non-empty, indistinguishable groups. Then we assign these 4 groups to the 4 distinct employees (which is \(4!\) ways).

We need to calculate \(S(5, 4)\). To partition 5 distinct jobs into 4 non-empty sets, the only possible structure is one set containing 2 jobs and the other three sets containing 1 job each (structure 2, 1, 1, 1). The number of ways to do this is by choosing which 2 jobs go together into the single set with more than one element.

So, there are 10 ways to partition the 5 jobs into 4 non-empty groups.

Now, we assign these 4 groups to the 4 distinct employees. Since the employees are distinct, there are \(4!\) ways to assign these 4 groups to the 4 employees.

The total number of ways to assign the 5 distinct jobs to the 4 distinct employees such that each employee gets at least one job is the product of the number of ways to partition the jobs and the number of ways to assign the partitions to the employees:

Both methods yield the same result.

Conclusion

There are 240 ways to assign 5 different jobs to 4 different employees such that every employee is assigned at least 1 job.

Revision Table: Key Concepts in Job Assignment Problems

Additional Information: Related Counting Problems

Combinatorics offers tools for various counting problems. Here are some related scenarios:

• Assigning different jobs to different employees without restrictions: As calculated, this is \(n^m\), where \(m\) is the number of jobs and \(n\) is the number of employees. Each of the \(m\) jobs has \(n\) choices.
• Assigning different jobs such that a specific employee gets no job: Choose the employee (\(\binom{n}{1}\)), then assign all jobs to the remaining \(n-1\) employees: \(\binom{n}{1}(n-1)^m\).
• Assigning identical jobs to different employees: This is a stars and bars problem. If there are \(m\) identical jobs and \(n\) different employees, the number of ways to assign them is \(\binom{m+n-1}{n-1}\).
• Assigning identical jobs to different employees with each employee getting at least one: Similar to the above, but we first give one job to each employee, leaving \(m-n\) jobs to distribute. The number of ways is \(\binom{(m-n)+n-1}{n-1} = \binom{m-1}{n-1}\).
• Assigning different jobs to identical machines: This is about partitioning a set of \(m\) distinct objects into \(n\) indistinguishable subsets, which is the definition of the Stirling number of the second kind, \(S(m, n)\). The problem requires each machine to get at least one job, so we need non-empty subsets.

Understanding the nature of the objects being assigned (distinct or identical) and the recipients (distinct or identical) is crucial in determining the correct counting method.