Problem 37
Question
Six performers are to present their comedy acts on a weekend evening at a comedy club. One of the performers insists on being the last stand-up comic of the evening. If this performer’s request is granted, how many different ways are there to schedule the appearances?
Step-by-Step Solution
Verified Answer
Thus, there are 120 different ways to schedule the appearances of the performers while honoring the last performer’s request.
1Step 1: Identify scenario and apply permutation formula
In this exercise of arranging the comic performers, repetition is not allowed. This is a typical case of permutations. The number of permutations can be found with the formula \( P(n,r) = \frac{n!}{(n-r)!} \), where \( n \) is the total number of objects or items and \( r \) is the number of items that are to be selected and arranged. Since one performer’s position is already fixed (at the end), this leaves us with 5 other performers to arrange.
2Step 2: Calculate permutation
With 5 performers to be arranged in 5 spots, use the permutation formula, resulting in \( P(5,5) = \frac{5!}{(5-5)!} \). That simplifies to \( \frac{5!}{0!} \). Because 0! is defined as 1, the result is simply \( 5! \).
3Step 3: Find factorial
Calculate \( 5! \) which is \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). This means there are 120 different possible arrangements for the 5 performers.
4Step 4: Consider the performer who insists on being the last
The 6th performer that insists on going last can only be in one position. Therefore, there's only one way to schedule the 6th performer. Multiply this 1 way for the 6th performer by the 120 ways for the first 5 performers.
Key Concepts
FactorialArrangementsCombinatorics
Factorial
In the world of mathematics, a factorial is a key concept used to define permutations and combinations. The factorial of a non-negative integer, denoted as
- The symbol \( n! \) signifies the multiplication of all positive integers up to \( n \). For instance, to find \( 5! \), you compute \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
- An important property of factorials is that \( 0! \) equals 1. This might seem strange, but it's defined this way to make the math work in situations like permutations and combinations.
Arrangements
Arrangements are a fundamental part of solving problems that involve organizing objects in specific sequences or orders. Each possible sequence is called an arrangement or permutation. In scenarios where the order matters, such as scheduling performances, arrangements are crucial.
Here are some basics:
- When solving for arrangements, ask whether the order of objects is important. This distinguishes permutations from combinations.
- If you need to arrange all items, like the 5 performers in our exercise, use permutations.
- The position of one or more objects can be fixed, as with the performer who insists on being last. This reduces the number of objects to be permuted.
Combinatorics
Combinatorics is a field in mathematics that deals with counting, arrangement, and combination of objects. It provides us with the tools to calculate permutations and combinations, which are crucial for organizing objects in certain orders.
When you're working with permutations, as in the comedy club example, you're concerned with the order of selection:
- Permutations are used when the arrangement or sequence of the items matters.
- In the given problem, the order of performers is significant, making this a combinatorics problem involving permutations.
Other exercises in this chapter
Problem 37
Write the first three terms in each binomial expansion, expressing the result in simplified form. $$ \left(y^{3}-1\right)^{20} $$
View solution Problem 37
find each indicated sum. $$ \sum_{i=5}^{9} 11 $$
View solution Problem 37
Find the sum of the first 50 terms of the arithmetic sequence: \(-10,-6,-2,2, \dots\)
View solution Problem 38
Find the sum of each infinite geometric series. $$ 1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots $$
View solution