Question

In Exercises 37-40, tell whether the question can be answered using permutations or combinations. Explain your reasoning. Then answer the question. To complete an exam, you must answer 8 questions from a list of 10 questions. In how many ways can you complete the exam?

Short answer

There are 45 ways to answer 8 questions out of a total of 10 questions for the exam.

Step-by-step solution

Identify n and k

We can see that we have a total of 10 questions (this is our \(n\)) from which we need to choose 8 (this is our \(k\)).

Apply the formula for combinations

We can plug our values for \(n\) and \(k\) into the combination formula to calculate the total number of ways to choose 8 questions from 10. So, \[C(10,8) = \frac{10!}{8!(10-8)!}\]

Calculate factorials and final result

Now we have to calculate the factorials and simplify the above equation. \[C(10,8) = \frac{10*9*8*7*6*5*4*3*2*1}{8*7*6*5*4*3*2*1*2*1}\]Simplifying above, we get \[ = \frac{10*9*8*7*6*5*4*3*2*1}{8*7*6*5*4*3*2*1*2*1} = \frac{10*9}{2*1} = 45\]

Key concepts

Permutations vs Combinations

Understanding the concepts of permutations and combinations is fundamental in the study of combinatorics and is crucial for solving various types of counting problems. Simply put, permutations are about arranging items in a specific order, whereas combinations are about selecting items without considering the order.

For example, consider a scenario where you have three different books. If we're interested in all the ways we can arrange these books on a shelf, we are dealing with permutations. However, if we simply want to select two out of three books to take on a trip, where the order doesn't matter, we are looking at combinations.

The primary difference between permutations and combinations lies in the importance of order. In permutations, 'ABC' is different from 'BCA'. But in combinations, selecting 'A' and 'B' is the same as selecting 'B' and 'A'. The respective formulas also reflect this difference. For permutations, we use nPr = \( \frac{n!}{(n-k)!} \), and for combinations, we use nCr = \( \frac{n!}{k!(n-k)!} \), where n is the total number of items and k is the number of items to choose or arrange.

In the context of the original exercise, since the order in which you answer the exam questions doesn't matter, we use combinations to find out how many ways you can choose 8 questions out of 10.

Factorials in Algebra

Factorials in algebra are a way to express the product of an integer and all the positive integers below it. It's denoted by an exclamation point (!). For instance, 5! (read '5 factorial') would be 5 x 4 x 3 x 2 x 1. Factorials are crucial in calculating permutations and combinations because they allow us to handle large quantities of multiplying sequential numbers.

One crucial property of factorials is that 0! is defined to be 1, which is an essential convention in many mathematical formulas. Understanding how to manipulate factorials is key when simplifying combination and permutation equations. For instance, in the step-by-step solution provided, simplifying \( \frac{10!}{8!(10-8)!} \) becomes much easier when you recognize that 10! (ten factorial) includes the product of 8! (eight factorial) within it.

So, instead of writing out all the terms and cancelling, we can shorten the process by cancelling out the common factorial terms immediately, recognizing that n! contains k! for any k less than n. This understanding is especially valuable as numbers grow larger, where directly calculating factorials becomes impractical.

Combinatorics

Combinatorics is the branch of mathematics devoted to counting, ordering, and arranging. It comes into play in a range of fields from computer science to statistics and plays a pivotal role in the study of probability and algebra.

Counting involves determining the number of ways certain events can occur, and this is where the importance of permutations and combinations is most evident. In an academic setting, combinatorics can help us tackle problems that ask, 'In how many ways can a particular event occur?' just like the exercise with the exam questions.

Combinatorial problems often require a deep understanding of factorials, permutations, and combinations, which are the building blocks of more complex combinatorial formulas and techniques such as Pascal's Triangle, Binomial Theorem, and the Principle of Inclusion and Exclusion. Each of these tools helps us to count in different scenarios, where direct counting strategies are inefficient or impractical.

It is essential to identify what the problem is asking to select the appropriate combinatorial tool. In the given exercise, because the order of answering the questions is unimportant, it was a combination problem, a concept regularly applied within combinatorics.