(a) There are 10 chairs in a row and 8 people to be seated. In how many ways can this be done? (b) There are 10 questions on a test and you are to do 8 of them. In how many ways can you choose them? (c) In part (a) what is the probability that the first two chairs in the row are vacant? (d) In part (b), what is the probability that you omit the first two problems in the test? (e) Explain why the answer to parts (a) and (b) are different, but the answers to (c) and (d) are the same.

Permutations are all about arranging objects in a specific order. When dealing with permutations, the order matters. For example, if we have a set of 8 people and 10 chairs, each distinct arrangement of people in the chairs is considered a unique permutation. The general formula to find the number of permutations, where we need to arrange r items out of a total of n items, is: \[ P(n, r) = \frac{n!}{(n-r)!} \]. This formula ensures that we consider every possible order of r items selected from n items. For instance, in our exercise, to find the number of ways to arrange 8 people in 10 chairs, we calculate \[ P(10, 8) = \frac{10!}{2!} = 1814400 \]. This large number shows just how many different ways 8 people can be seated in 10 chairs!

Combinations are about selecting items where the order does not matter. Unlike permutations, here it only matters which items are chosen, not their arrangement. The formula for combinations to choose r items from a set of n items is given by: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]. In our exercise, when choosing 8 questions to answer out of 10, the order of answering doesn’t matter. Hence, we use the combination formula:\[ C(10, 8) = \frac{10!}{8!2!} = 45 \].This result tells us that there are 45 different ways to select 8 questions out of 10.

Probability calculations help us determine the likelihood of a particular event occurring. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For instance, in part (c) of our exercise, we calculate the probability that the first two chairs are vacant when arranging 8 people out of 10 chairs. We find the number of favorable outcomes: \[ P(8, 8) = 8! = 40320 \]. Dividing this by the total number of arrangements calculated earlier: \[ \text{Probability} = \frac{8!}{P(10,8)} = \frac{40320}{1814400} = \frac{1}{45} \]. Similarly, in part (d) where we calculate the probability of omitting the first two problems out of the 10 questions, since there is only one way to choose (omit both), our favorable outcomes are: \[ C(8, 8) = 1 \]. Hence, the probability: \[ \text{Probability} = \frac{C(8, 8)}{C(10, 8)} = \frac{1}{45} \].Both parts (c) and (d) result in a probability of \( \frac{1}{45} \).

Factorials are mathematical expressions written as \( n! \) which represent the product of all positive integers up to n. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). They are crucial in calculating permutations and combinations because they simplify counting large sets of items. In permutations, for arranging r items out of n, we often see terms like \( \frac{n!}{(n-r)!} \), and in combinations, we use expressions like: \( \frac{n!}{r!(n-r)!} \).Let's break it down with part (a) from our exercise involving 10 chairs and 8 people. We used: \( P(10, 8) = \frac{10!}{2!} \). To calculate this, we first determine \( 10! \): \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800\), and \( 2! = 2 \times 1 = 2 \). Thus, \( \frac{10!}{2!} = \frac{3628800}{2} = 1814400 \). This result underlines the importance of understanding and accurately calculating factorials in probability and combinatorics.