Problem 62 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A student taking an examination is required to answer exactly 10 out of 15 questions.

Step-by-Step Solution

Answer

There are 3,003 different ways the student can choose 10 questions to answer out of 15 using the combination formula: \(C(15, 10) = \frac{15!}{10!5!} = 3003\).

Step 1: Understand the concept of combinations

A combination is a selection of items from a larger set, where the order of the items does not matter. The number of combinations of n items taken r at a time is denoted by C(n, r), which can be calculated using the formula: \[C(n, r) = \frac{n!}{r!(n-r)!}\] where n! (called "n factorial") is the product of all positive integers up to n, and is defined as: \[n! = n*(n-1)*(n-2)* \dots *2*1\]

Step 2: Identify the given values in the problem

In this problem, there are 15 questions (n = 15) and the student must choose 10 questions to answer (r = 10). We want to calculate the number of possible combinations of choosing 10 questions out of 15.

Step 3: Calculate the factorials

We need to calculate the factorials for n, r, and n-r: 1. 15! (n!): \(15*14*13*12*11*10*9*8*7*6*5*4*3*2*1\) 2. 10! (r!): \(10*9*8*7*6*5*4*3*2*1\) 3. 5! ((n-r)!): \(5*4*3*2*1\)

Step 4: Apply the combination formula

Now, we can apply the formula for counting combinations: \[C(n, r) = \frac{n!}{r!(n-r)!}\] Plug in the values for n, r, and their respective factorials: \[C(15, 10) = \frac{15!}{10!5!}\]

Step 5: Simplify the expression and find the number of combinations

Simplify the expression by dividing out the factorials: \[C(15, 10) = \frac{15*14*13*12*11*10*9*8*7*6*5*4*3*2*1}{(10*9*8*7*6*5*4*3*2*1)(5*4*3*2*1)}\] Cancel out the common terms in the numerator and denominator: \[C(15, 10) = \frac{15*14*13*12*11}{5*4*3*2*1}\] Calculate the result: \[C(15, 10) = \frac{360360}{120}\] \[C(15, 10) = 3003\] There are 3,003 different ways the student can choose 10 questions to answer out of 15.

Key Concepts

Factorial Notation

When we talk about factorial notation in mathematics, we refer to a function that plays a crucial role, particularly in combinatorial problems. Factorial notation is symbolized by an exclamation point (!) and calculated for a non-negative integer, n, as the product of all positive integers less than or equal to n. The function is defined as:

\[\[\begin{align*} n! = n \times (n-1) \times (n-2) \times \ ... \times 3 \times 2 \times 1.\end{align*}\]\]For instance, when calculating 5!, we multiply all positive integers from 5 down to 1:\[\[\begin{align*} 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120.\end{align*}\]\]A special case to remember is 0!, which is always equal to 1. This is a convention in mathematics to help simplify expressions involving factorials.
In the exercise provided, understanding factorials is integral for calculating the number of combinations. This is because the formula for combinations includes factorial terms that require simplification, as in the step where 15!, 10!, and 5! are calculated and then used to find the number of ways to choose 10 questions out of 15.

Permutations and Combinations

Permutations and combinations are both concepts used in counting problems within mathematics, with a key distinction between the two. Permutations involve arrangements where the order matters, while combinations involve selections where the order does not matter.

With permutations, you calculate the number of ways to arrange a certain number of items. For example, the number of permutations of three objects a, b, and c taken three at a time is 6, represented by the different orderings (abc, acb, bac, bca, cab, cba).

Combination Formula

For combinations, the order of selection is irrelevant, and the formula for calculating combinations is:
\[C(n, r) = \frac{n!}{r!(n-r)!}\]
Here, n is the total number of items, and r is the number of items being chosen. In the textbook problem, the students were required to select 10 out of 15 questions to answer, which directly employs the concept of combinations, as the arrangement of the chosen questions is not a concern.
Giving students clear examples of permutations and combinations helps to solidify their understanding. For instance, choosing a president, vice-president, and secretary from a group of 10 yields a permutation count, whereas simply selecting 3 committee members from the same group is a combination scenario.

Applied Mathematics

Applied mathematics is a branch of mathematics that concerns itself with mathematical methods that are typically used in science, engineering, business, and industry. Thus, 'applied mathematics' is a blend of specialized knowledge and mathematical science. A key aspect of applied mathematics is the use of mathematical models to solve real-world problems through concrete formulations and quantifiable predictions.

In our textbook exercise, the concept of combinations is a pure mathematical idea being applied to solve a practical question - how a student can strategize answering exam questions. This illustrates that combinatorial analysis is fundamentally an applied mathematics problem, which can help in diverse areas such as operations research, probability theory, and planning and optimization scenarios.
Educating students in how these abstract concepts manifest in tangible problems ensures that they recognize the relevance of mathematics in everyday life and decision-making processes. As seen with the textbook problem, students could apply this knowledge to maximize their potential exam scores by understanding the combinatorial possibilities.