Problem 26
Question
A mathematics department has ten faculty members but only nine offices, so one office must be shared by two individuals. In how many different ways can the offices be assigned?
Step-by-Step Solution
Verified Answer
There are 16,329,600 different ways to assign the offices.
1Step 1: Calculate Total Office Arrangements Without Restrictions
First, calculate the total number of ways to assign each of the 10 faculty members to 9 offices assuming no restrictions. This would simply be a permutation, since each faculty member can go into a different office. So, the number of ways is given by finding how to arrange 10 members in 9 offices:\[ 10^9 \]
2Step 2: Identify the Unique Arrangement Due to Shared Office
Since one office must be shared by two individuals, we need to choose two faculty members to share one office. The number of ways to pick 2 people from 10 to share an office is given by combinations:\[ \binom{10}{2} \]
3Step 3: Calculate Total Permutations for Nine Assignments
After selecting the 2 people to share one office, you will have 9 remaining people who need to be placed into the remaining 8 vacancies. So, the number of permutations for assigning 9 faculty members to 8 offices is calculated by:\[ 9! \]
4Step 4: Combine Combinatorial Choices and Permutations
Multiply the number from Step 2 by the result from Step 3 to get the total number of different ways to assign offices:\[ \binom{10}{2} \times 9! \]
5Step 5: Perform the Calculations
Calculate the combination and permutation:\[ \binom{10}{2} = 45 \]\[ 9! = 362880 \]Multiply them together:\[ 45 \times 362880 = 16329600 \]
Key Concepts
PermutationsCombinationsFaculty Office AssignmentMathematical Problem Solving
Permutations
In combinatorics, permutations refer to the different arrangements of a set of items where the order is crucial. For instance, when assigning offices, each distinct arrangement counts as a unique permutation. The formula for permutations of arranging a set of items is given by the factorial notation. When arranging 9 faculty members in 8 offices, we calculate it using the factorial of 9, written as \(9!\). This notation signifies the multiplication of all positive integers up to 9, resulting in a total of 362,880 possible arrangements. Permutations help determine the total number of varying sequences possible when distributing faculty members to offices.
Combinations
Combinations, in contrast to permutations, focus on selecting items where the order does not matter. This concept is relevant when choosing which two faculty members will share an office in the department. The number of ways to select 2 members from a group of 10 involves using the combination formula, denoted as \(\binom{n}{k}\) or 'n choose k'. Here, it reflects the number of ways to select k items from n items without regard to order. Using the formula \(\binom{10}{2}\), we find that there are 45 ways to choose which two faculty members should share an office.
Faculty Office Assignment
In a scenario like this, the department faces a common logistical problem: not enough offices for faculty members. With fewer offices than members, one office needs to be shared. The assignment process entails two key decisions:
- Selecting which two faculty members will share an office.
- Arranging the remaining faculty in the other offices.
Mathematical Problem Solving
Tackling a problem like the faculty office assignment involves several mathematical strategies. It starts by breaking down the challenge into manageable parts and determining how many ways tasks can be completed. The blend of permutations and combinations aids in finding the solution. Here are the steps usually taken in mathematical problem solving:
- Understand the problem: Clearly define what needs to be solved; in this case, assigning offices.
- Plan a strategy: Think about using combinations to select and permutations to arrange.
- Execute the plan: Perform the calculations using factorials and combinations.
- Review the solution: Verify calculations to ensure correctness.
Other exercises in this chapter
Problem 25
True-or-false test In how many different ways can a test consisting of ten true-or-false questions be completed?
View solution Problem 25
Exer. 23-28: Find the sum. $$ \sum_{k=1}^{18}\left(\frac{1}{2} k+7\right) $$
View solution Problem 26
Use the binomial theorem to expand and simplify. $$ 5\left(\frac{1}{2} x+y^{3}\right)^{4} $$
View solution Problem 26
Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ a+b \text { is a factor of } a^{2 n-1}+b^{2 n-1} $$
View solution