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Show Programs Three bands and two comics are performing for a student talent show. How many different programs (in terms of order) can be arranged? How many if the comics must perform between bands?

Short Answer

Expert verified
120 programs total; 12 if comics perform between bands.

Step by step solution

01

Identifying the Total Performing Acts

There are three bands and two comics, making a total of 5 acts. We need to consider how we can arrange these acts in different orders.
02

Calculating Total Arrangements

To find the total number of programs possible, we calculate the permutations of 5 acts. The formula for permutations is:\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \].This means there are 120 different ways to arrange the programs.
03

Setting Condition for Comics Performing Between Bands

Now, let’s address the condition where comics must perform between the bands. Order the bands first and consider slots between them for the comics.
04

Arranging Bands with Slots for Comics

Arrange the 3 bands in sequence, creating 4 slots where the comics can potentially perform: *(before the first band, between bands, and after last band). With 3 bands, the arrangement is 3! = 6. The slots will look like: - B - B - B -.
05

Filling Slots with Comics

We want comics to appear only in the spaces between the bands. Therefore, the comics need to go in the 2 slots *between* the bands. Since there are 2 comics, we can place them in the slots as follows: \[ 2! = 2 \times 1 = 2 \].Therefore, the 2 comics can be arranged in these slots in 2 different ways.
06

Total Arrangements with Condition

Multiply the arrangements of the bands by the arrangements of the comics to get the total number of arrangements where the comics perform only between the bands:\[ 6 \text{(bands)} \times 2 \text{(comics)} = 12 \].This means there are 12 different arrangements where the comics perform between bands.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutations
Permutations are an essential concept in combinatorics, often used when calculating the number of possible arrangements of a set of items. In many cases, the order of arrangement is significant, so each variation of order is counted as a separate permutation.
For example, when arranging different acts in a talent show, like the three bands and two comics in our scenario, each different order of appearance constitutes a distinct permutation. The general formula for permutations of a set of n different items is given by the factorial function, denoted as\[ n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1 \].
This formula helps us determine how to arrange all items. Understanding permutations is crucial for solving various problems where order matters, making it a fundamental tool in statistics and probability.
Factorial Calculation
Factorial forms the heart of permutation calculations. It represents the product of all positive integers up to a specified number, , and is symbolized by "!". For instance, for the number 5, the factorial, denoted as 5!, is\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \].
Factorials grow rapidly with increasing numbers, and they are essential in computing permutations and combinations in statistics. Every permutation or combination problem starts by determining how many items we are arranging and then calculating the total length of the arrangement using factorials.
The concept of factorial calculation simplifies the process of determining different possible arrangements or selections, especially when the datasets are relatively small with distinct elements to organize.
Conditional Arrangements
Conditional arrangements refer to permutations that must satisfy specific conditions or rules. In many scenarios, like our talent show, certain orders or types of elements must be arranged in a specific manner relative to others.
For instance, given that comics must perform only "between" bands, we first arrange the bands as desired and then determine appropriate slots for the comics.
To achieve this:
  • First, arrange the 3 bands using their permutations\[ 3! = 6 \], resulting in different band sequences.
  • Next, identify possible comic slots, where performers like the comics are inserted into the defined spaces \( '-' \) between the bands.
  • Finally, arrange the comics in these specific slots\[ 2! = 2 \], yielding confirmed placements for the conditions.
Combining these steps yields the total number of viable arrangements fitting the given criteria. Conditional arrangements enhance the understanding of how specific conditions influence potential outcomes in statistical calculations.

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Most popular questions from this chapter

Tossing a Coin and Rolling a Die A coin is tossed; if it falls heads up, it is tossed again. If it falls tails up, a die is rolled. Draw a tree diagram and determine the outcomes.

$$ \begin{array}{l}{\text { Odds Odds are used in gambling games to make them }} \\\ {\text { fair. For example, if you rolled a die and won every }} \\\ {\text { time you rolled a } 6, \text { then you would win on average }} \\\ {\text { once every } 6 \text { times. So that the game is fair, the odds of }} \\ {5 \text { to } 1 \text { are given. This means that if you bet } \$ 1 \text { and won, }} \\ {\text { you could win } \$ 5 . \text { On average, you would win } \$ 5 \text { once }} \\ {\text { in } 6 \text { rolls and lose } \$ 1 \text { on the other } 5 \text { rolls-hence the }} \\ {\text { term fair game. }}\end{array} $$ $$ \begin{array}{l}{\text { In most gambling games, the odds given are not fair. }} \\ {\text { For example, if the odds of winning are really } 20 \text { to } 1,} \\ {\text { the house might offer } 15 \text { to } 1 \text { in order to make a profit. }} \\ {\text { Odds can be expressed as a fraction or as a ratio, }} \\ {\text { such as } \frac{5}{1}, 5: 1, \text { or } 5 \text { to } 1 . \text { Odds are computed in favor }} \\ {\text { of the event or against the event. The formulas for }} \\ {\text { odds are }}\end{array} $$ $$ \begin{array}{l}{\text { Odds in favor }=\frac{P(E)}{1-P(E)}} \\ {\text { Odds against }=\frac{P(\bar{E})}{1-P(\bar{E})}}\end{array} $$ In the die example, $$ \begin{array}{c}{\text { Odds in favor of a } 6=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5} \text { or } 1: 5} \\ {\text { Odds against a } 6=\frac{\frac{5}{6}}{\frac{1}{6}}=\frac{5}{1} \text { or } 5: 1}\end{array} $$ Find the odds in favor of and against each event. a. Rolling a die and getting a 2 b. Rolling a die and getting an even number c. Drawing a card from a deck and getting a spade d. Drawing a card and getting a red card e. Drawing a card and getting a queen f. Tossing two coins and getting two tails g. Tossing two coins and getting exactly one tail

Rolling Two Dice If two dice are rolled one time, find the probability of getting these results: a. A sum of 5 b. A sum of 9 or 10 c. Doubles

Crimes Committed The numbers show the number of crimes committed in a large city. If a crime is selected at random, find the probability that it is a motor vehicle theft. What is the probability that it is not an assault? $$ \begin{array}{ll}{\text { Theft }} & {1375} \\ {\text { Burglary of home or office }} & {500} \\ {\text { Motor vehicle theft }} & {275} \\ {\text { Assault }} & {200} \\ {\text { Robbery }} & {125} \\ {\text { Rape or homicide }} & {25}\end{array} $$

When an event is certain to occur, what is its probability?

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