(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 fint 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 answers to parts (a) and (b) are different, but the answers to (c) and (d) are the same.

Permutations are fundamental in combinatorics. They describe the different ways of ordering a set of objects where order matters. For example, if you have 10 chairs and want to seat 8 people, you will use permutations to figure out how many possible seating arrangements you can have.

The formula for permutations is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \]
Here, \(!\) denotes a factorial, meaning you multiply the series of descending natural numbers. In practical terms, this means 10! (10 factorial) is 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.

In part (a) of the exercise, to find the number of ways to seat 8 people in 10 chairs, we compute: \[ P(10, 8) = \frac{10!}{(10-8)!} = \frac{10!}{2!} = \frac{10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}{2 × 1} \] This calculation gives us a very large number of possible arrangements.

Combinations, unlike permutations, are used when the order does not matter. For instance, if you need to choose 8 questions out of 10 on a test, you’ll use combinations to determine the number of ways to select the questions.

The formula for combinations is: \[ C(n, r) = \frac{n!}{r! (n-r)!} \]
This accounts for selecting 8 out of 10 questions without considering their order.

For part (b) of the exercise, calculate: \[ C(10, 8) = \frac{10!}{8!2!} = \frac{10 × 9}{2 × 1} = 45 \] This tells you there are 45 different ways to choose 8 questions out of 10 without worrying about the order in which you choose them.

Probability measures the likelihood of an event happening. It is calculated by dividing the number of favorable outcomes by the number of possible outcomes.

In part (c) of the exercise, if the first two chairs are vacant, we need to calculate the number of favorable seating arrangements: \[ P(8, 8) = 8! \]
The total ways to seat 8 people in 10 chairs are: \[ P(10, 8) = \frac{10!}{2!} \]
Therefore, the probability of having the first two chairs vacant is: \[ \frac{8!}{\frac{10!}{2!}} = \frac{8! × 2!}{10!} = \frac{1}{45} \]
This tells us there is a 1 in 45 chance that the first two chairs will be vacant.

Similarly, for part (d), we find the probability of not choosing the first two problems on the test:

First, calculate the number of ways to choose 8 out of the remaining 8 questions, excluding the two: \[ C(8, 8) = 1 \]
The total ways to select 8 questions out of 10: \[ C(10, 8) = 45 \]
Hence, the probability is: \[ \frac{1}{45} \] This shows the same probability, reinforcing the concept.

Factorials are a key part of combinatorial calculations. They represent the product of an integer and all the integers below it.

The notation n! (read as 'n factorial') means: \[ n! = n × (n-1) × (n-2) × ... × 1 \]
Factorials are used in both permutations and combinations.

For example, 10! (10 factorial) is:
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800.

Factorials grow very rapidly in size, which is why it's often easier to think about them in pieces rather than directly calculate large values.

In permutations and combinations, factorials help reduce the complexity of the calculations. Understanding how to manipulate and simplify factorials is crucial for solving combinatorial problems efficiently.