Problem 14 - Algebra for College Students

Question

In how many ways can a history professor randomly assign exactly one \(\mathrm{A},\) one \(\mathrm{B},\) one \(\mathrm{C},\) one \(\mathrm{D},\) and \(\operatorname{six} \mathrm{F}^{\prime}\) s to a class of 10 students?

Step-by-Step Solution

Answer

5040 ways

Step 1: Understand the Problem

Identify that we need to find the number of ways to assign different grades (A, B, C, D, and six F's) to 10 students. The grades are distinct except for the F's, which are identical.

Step 2: Total Number of Students and Grades

There are 10 students and 10 grades (one of each grade A, B, C, D, and six F's).

Step 3: Calculate the Permutations

We use the formula for permutations of a multiset to calculate the total number of ways to assign the grades: \[ \frac{10!}{6!1!1!1!1!} = \frac{10!}{6!} \]

Step 4: Simplify the Factorial Expression

Simplify the factorial expression: \[ \frac{10!}{6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} = 10 \times 9 \times 8 \times 7 = 5040 \]

Key Concepts

Factorials

Factorials are a fundamental concept in mathematics, especially in counting problems and permutations. The factorial of a number, denoted as \( n! \), is the product of all positive integers less than or equal to that number. For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials grow very quickly with larger numbers. In our history professor problem, we use \( 10! \) and \( 6! \). Here, \( 10! \) is the factorial of 10, which equals \( 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \). For comparison, \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \). By understanding how to manipulate factorials, we can simplify complex permutations as seen in our solution. This simplification helps to find the number of ways to assign grades accurately.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations, permutations, and counting. It enables us to solve problems about ordering and selecting objects. In the professor's grade assignment scenario, combinatorics helps determine the number of ways to assign grades. Since grades A, B, C, and D are unique but there are six F's, we use the permutation formula for a multiset. A multiset allows for repetition of some elements, unlike a simple set.
The formula for permutations of a multiset is given by: \[ \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \]
Here, \( n \) is the total number of items to arrange (10 students), and the denominators \( n_1! \) to \( n_k! \) are the factorials of the counts of each distinct item. In our case, it translates to: \[ \frac{10!}{6! \times 1! \times 1! \times 1! \times 1!} = \frac{10!}{6!} \] This approach simplifies the counting process by accounting for the repeated F grades effectively.

Grade Assignment Problem

The grade assignment problem is a classic example of how combinatorics and factorials come together to solve practical issues. To determine the number of ways to assign grades such as A, B, C, D, and six F's to 10 students, we must consider both permutation and repetition.
Here’s how it works step-by-step:

• Identify that there are 10 grades for 10 students.
• Recognize that four grades (A, B, C, D) are unique, but there are six identical F's.
• Apply the permutation formula for a multiset to determine the number of unique ways to distribute these grades.

Simplifying the expression, we get:
\[ \frac{10!}{6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} = 10 \times 9 \times 8 \times 7 = 5040 \] Thus, the history professor has 5040 unique ways to assign these grades. This method ensures that all possible arrangements are considered without over-counting similar configurations of repeating grades.