Problem 63 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

TEACHING AssISTANTSHIPS Twelve graduate students have applied for three available teaching assistantships. In how many ways can the assistantships be awarded among these applicants if a. No preference is given to any student? b. One particular student must be awarded an assistantship? c. The group of applicants includes seven men and five women and it is stipulated that at least one woman must be awarded an assistantship?

Step-by-Step Solution

Answer

In summary: a. There are 220 ways to award the assistantships without any preference. b. There are 55 ways to award the assistantships, ensuring one particular student gets an assistantship. c. There are 185 ways to award the assistantships, ensuring at least one woman gets one.

Step 1: Find the number of ways the assistantships can be awarded

We have 12 applicants and 3 assistantships to be awarded. We will use the combinations formula to find the number of ways which is given as: nCr = n! / (r!(n-r)!) In this case, n = number of students (12) and r = number of assistant positions (3). Therefore: 12C3 = \( \frac{12!}{(3!(12-3)!)} \)

Step 2: Calculate the combinations

Now we will calculate the value for 12C3: 12C3 = \( \frac{12!}{(3!(9)!)} \) = \( \frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{(3\times2\times1)(9\times8\times7\times6\times5\times4\times3\times2\times1)} \) = 220 Thus, 220 different ways exist to distribute the 3 assistantships among the applicants without any preference. #b. One particular student must be awarded an assistantship#

Step 1: Number of remaining students and assistant positions

If one particular student must get one assistantship, then we have 11 remaining students and 2 assistant positions.

Step 2: Calculate the combinations for the remaining spots

Using the combinations formula, we have: 11C2 = \( \frac{11!}{(2!(11-2)!)} \) Now we will calculate the value for 11C2: 11C2 = \( \frac{11!}{(2!(9)!)} \) = \( \frac{11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{(2\times1)(9\times8\times7\times6\times5\times4\times3\times2\times1)} \) = 55 Therefore, there are 55 ways of awarding the assistantships, ensuring one particular student gets an assistantship. #c. At least one woman must be awarded an assistantship#

Step 1: Calculate the total ways without restrictions

First, find the total number of ways (which we already solved in part a) without any restrictions. As a reminder: 12C3 = 220

Step 2: Calculate the combinations for all men getting an assistantship

Next, we calculate the combinations where all 3 assistantships go to the 7 men, leaving the women out: 7C3 = \( \frac{7!}{(3!(7-3)!)} \) = \( \frac{7\times6\times5\times4\times3\times2}{(3\times2\times1)(4\times3\times2\times1)} \) = 35

Step 3: Calculate the number of ways with at least one woman

Now we will subtract the outcomes where all men got the assistantships from the total combinations: Ways with at least one woman = Total ways - Ways with all men Ways with at least one woman = (220 - 35) = 185 Thus, there are 185 different ways to award the assistantships, ensuring at least one woman gets one.

Key Concepts

Permutations and Combinations

Understanding the basic principles of permutations and combinations is essential in solving many problems related to probability and combinatorics. When we consider permutations, we focus on the arrangement of objects where the order matters. For instance, the sequence ABC is different from BAC. On the other hand, combinations deal with the selection of objects where the order doesn't matter, hence ABC and BAC are considered the same.
In education, especially when discussing the selection of candidates for roles or positions, we employ combinations as we are typically interested in the group chosen, not the order they were selected in. The formula for combinations is
\(nCr = \frac{n!}{r!(n-r)!}\), where 'n' is the total number of items to choose from, 'r' is the number of items to choose, and '!' denotes a factorial. In our teaching assistantship exercise, combinations are used to determine the number of ways to select 3 assistants from 12 applicants, not caring about the order they are chosen.

Factorial

In mathematics, the factorial of a non-negative integer is the product of all positive integers less than or equal to that integer. It's represented by an exclamation point (!). For example, the factorial of 5 is written as \(5!\) and is calculated as \(5 \times 4 \times 3 \times 2 \times 1\), which equals 120.
Factorials are fundamental in calculating permutations and combinations because they enable us to count the number of ways in which items can be arranged or selected. The concept of factorial is embedded in the formula for combinations used in our example problem, making it a cornerstone operation in combinatorial exercises. For those learning about combinatorics, understanding factorials is a vital step in mastering the subject.

Teaching Assistantships Selection

In academic institutions, teaching assistantships are often awarded based on certain criteria, and understanding the selection process can involve combinatorics. Our exercise addresses different scenarios in awarding three assistantships among twelve candidates. In scenario a, with no preferences, we use basic combinations to find there are 220 ways to award these positions. Introducing a preference, as in scenario b, where one specific student must be selected, reduces our pool and changes the calculation, resulting in 55 ways. The third scenario requires the inclusion of at least one woman in the selection, reflecting real-world conditions where diversity criteria might come into play.
In this case, the total possibilities without restrictions (220) are adjusted by removing arrangements that do not meet the criteria (in this case, all male selections, which amount to 35). This results in 185 acceptable combinations. These examples show how combinatorics can be applied to real-life situations, such as the fair and equitable distribution of academic roles.