Problem 64 - A Graphical Approach to College Algebra

Question

Use any or all of the methods described in this section to solve each problem. Assistant/Manager Assignments From a pool of 7 assistants, 3 are selected to be assigned to 3 managers, 1 assistant to each manager. In how many ways can this be done?

Step-by-Step Solution

Answer

210 ways to assign 3 assistants from 7 to 3 managers.

Step 1: Identify the Problem Type

This is a combination problem where we are trying to determine the number of ways to select 3 assistants from a pool of 7.

Step 2: Calculate the Number of Ways to Select Assistants

First, calculate the number of ways to choose 3 assistants from 7. This can be computed using combinations, defined as \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).For this problem, \( n = 7 \) and \( r = 3 \). Thus:\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]There are 35 ways to select 3 assistants from 7.

Step 3: Assign Assistants to Managers

After selecting 3 assistants, each can be assigned to one of the 3 managers. This is a permutation problem because the order of assignment matters.The number of permutations of 3 assistants is calculated as follows:\( 3! = 3 \times 2 \times 1 = 6 \)There are 6 possible ways to assign these assistants to the 3 managers.

Step 4: Total Number of Ways

To find the total number of ways to select and assign the assistants, multiply the number of combinations (ways to select assistants) by the number of permutations (ways to assign them to the managers):\[ 35 \times 6 = 210 \]Thus, there are 210 ways to select and assign 3 assistants to 3 managers.

Key Concepts

Permutations

Permutations are all about arranging items in a specific order. When the sequence of items matters, you are dealing with permutations. For instance, if you're planning to assign assistants to managers, the order in which each assistant is assigned plays a critical role.

For the given problem, after selecting 3 assistants, the real challenge is to arrange them so that each one is assigned to a different manager. Here, the order is vital because assigning Assistant A to Manager 1 and Assistant B to Manager 2 is different from assigning Assistant B to Manager 1 and Assistant A to Manager 2.

The number of possible arrangements or permutations of the 3 selected items (assistants) is calculated using the factorial of the number of items. In this case, it is calculated as follows:

• For 3 assistants, the permutations are given by the factorial: 3!, which equals 6.

Thus, there are 6 ways to assign the selected assistants to managers.

Combinations

Combinations focus on choosing a subset of items from a larger set, where the order of selection does not matter. This means that selecting items in different sequences counts as the same choice. Think of buying ice cream—whether you choose chocolate, vanilla, and then strawberry, or strawberry, vanilla, and chocolate, you've got the same combination of flavors.

In the exercise, our task is to choose 3 assistants from a group of 7. Here, the sequence of picking the assistants doesn't affect this choice, making it a combination problem. The formula to determine combinations is:

• \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Applying this to the problem, where there are 7 assistants and we need to choose 3:

• \( \binom{7}{3} = \frac{7!}{3!(7-3)!} \)
• Calculate further to find there are 35 ways to pick the needed assistants.

Factorial

The concept of factorial is crucial in both permutations and combinations. The factorial of a number, represented as \(!\), means multiplying that number by every whole number less than it down to 1. So, for any positive integer \( n \), \( n! \) stands for \( n \times (n-1) \times (n-2) \times ... \times 1 \).

The factorial concept is used in calculating both combinations and permutations because it helps account for the various ways you can arrange or select items.

Let's break it down further using examples:

• Example for combinations: In our problem, \( 7! \) helps to see all potential arrangements, while dividing by \( (7-3)! \) and \( 3! \) adjusts for over-counting since the order doesn't matter when choosing assistants.
• Example for permutations: For assigning to managers, we calculate \( 3! \) because each assistant is given a precise order of assignment, reflecting various arrangements.

Factorials simplify calculations in combinatorics by providing a structured way to look at arrangements and selections.