Question

(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.

Short answer

(a) 1814400, (b) 45, (c) 1/45, (d) 1/45. Permutation differs from combination, but the probabilities remain the same.

Step-by-step solution

Understand Part (a)

We need to find the number of ways to arrange 8 people in 10 chairs. Since each arrangement of people in chairs counts as a different way, we use the formula for permutations.

Calculate Permutations for 8 out of 10 Chairs

The number of permutations (arrangements) of 8 people in 10 chairs is given by the formula for permutations of n items taken r at a time: \[ P(n, r) = \frac{n!}{(n-r)!} \]Substitute n = 10 and r = 8: \[ P(10, 8) = \frac{10!}{(10-8)!} = \frac{10!}{2!} \]Calculate it: \[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]\[ 2! = 2 \times 1 \]\[ \frac{10!}{2!} = \frac{3628800}{2} = 1814400 \]

Understand Part (b)

We need to choose 8 questions out of 10. Since the order does not matter, we use the combination formula.

Calculate Combinations for Choosing 8 out of 10 Questions

The number of combinations of n items taken r at a time is given by: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]Substitute n = 10 and r = 8: \[ C(10, 8) = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} \]Calculate it:

Calculate Factorials

\[ 10! = 10 \times 9 \times 8! \]\[ \frac{10!}{8!2!} = \frac{10 \times 9 \times 8!}{8! \times 2 \times 1} = \frac{10 \times 9}{2} = 45 \]

Understand Part (c)

We need the probability that the first 2 chairs are vacant in the arrangement found in part (a).

Calculate Probability for First 2 Chairs Vacant

If the first two chairs are to be vacant, we need to arrange 8 people in the remaining 8 chairs:\[ P(8, 8) = 8! = 40320 \]To find the probability, divide the favorable outcomes by the total outcomes: \[ \frac{8!}{P(10, 8)} = \frac{40320}{1814400} = \frac{1}{45} \]

Understand Part (d)

We need the probability of omitting the first two problems from the test given the combination found in part (b).

Calculate Probability for Omitting First 2 Problems

If the first 2 problems are omitted, we need to choose 8 from the remaining 8 questions:\[ C(8, 8) = 1 \]To find the probability, divide the favorable outcomes by the total outcomes: \[ \frac{C(8, 8)}{C(10, 8)} = \frac{1}{45} \]

Explain the Differences in Parts (a) & (b) and Similarities in Parts (c) & (d)

Part (a) deals with permutations where order matters, and part (b) deals with combinations where order does not matter. Therefore, the number of ways to arrange and choose are different. Parts (c) and (d) result in the same probability because in both cases, we are selecting from the remaining items in an unordered fashion.

Key concepts

Permutations

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

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

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

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.