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The call letters of a radio station must have 4 letters. The first letter must be a K or a W. How many different station call letters can be made if repetitions are not allowed? If repetitions are allowed?

Short Answer

Expert verified
Without repetitions: 24,288. With repetitions: 35,152.

Step by step solution

01

Determine the First Letter

Since the first letter must be either 'K' or 'W', we have 2 possible choices for the first position of the call letters.
02

Calculate Options Without Repetition

For the remaining 3 letters, choose from the 24 remaining letters of the alphabet, because no repetitions are allowed. So, there are 24 choices for the second letter, 23 for the third, and 22 for the fourth letter. Calculate the total combinations: \[ 2 \times 24 \times 23 \times 22 = 24,288 \]
03

Calculate Options With Repetition

If repetitions are allowed, each of the remaining 3 positions has 26 possible choices (since the letter used once can be used again). Calculate the total combinations: \[ 2 \times 26 \times 26 \times 26 = 35,152 \]
04

Compare Both Scenarios

We find there are 24,288 possible combinations without repetitions and 35,152 possible combinations with repetitions. This comparison shows a significant increase in possible call letters when repetitions are allowed.

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Key Concepts

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

Permutations Without Repetition
When dealing with permutations without repetition, each item can only be selected once. After it has been chosen, it can't reappear in the same sequence. Consider having a distinct selection of people who can perform at your event. Once a person is chosen, they cannot be selected again for that particular performance lineup. This is akin to our radio station example, where no letter can be reused.

The key idea is that choices decrease with each selection:
  • The first letter of the call sign is either 'K' or 'W', giving us 2 options for the first position.
  • Next, since we're avoiding repetition, the second letter has only 24 remaining options.
  • The available letters shrink further to 23 choices for the third letter, followed by 22 for the fourth.

The total number of permutations without repetition can be calculated as: \[2 \times 24 \times 23 \times 22 = 24,288\]

This approach highlights how each choice reduces the pool of available options, making it crucial to prioritize and sequence them effectively.
Permutations With Repetition
In contrast to permutations without repetition, permutations with repetition allow for items to be reused in each position. This provides vastly more combinations. Imagine a digital password where each character can be any letter or number, even if previously used. The possibilities expand significantly compared to when each character must be unique.

For our radio station call letters when repetition is allowed:
  • The first letter, as before, can be either 'K' or 'W', providing 2 choices.
  • However, every following letter is selected from all 26 letters of the alphabet, because repetitions are allowed.
  • This means the second, third, and fourth letters each have 26 possible choices.


Thus, the total number of permutations with repetition is calculated as: \[2 \times 26 \times 26 \times 26 = 35,152\]

This method highlights how allowing the reuse of elements can exponentially increase the number of possible combinations.
Probability and Counting Techniques
Counting and probability are vital tools in combinatorial mathematics, helping us understand and organize different potential outcomes. They provide clarity on how many ways a scenario might unfold, which is particularly useful in situations involving uncertainty or large datasets.

Counting Techniques
In our radio station example, we used counting techniques to establish the number of possible call letters by methodically evaluating each position's options, both with and without repetition.
  • By understanding the context and conditions, such as "without repetition," we ensured calculations only counted unique sequences.
  • "With repetition," involved calculating when each option could appear multiple times throughout the arrangement.


Connection with Probability
Understanding permutations aids in calculating probabilities by showing the total number of possible outcomes. For instance, knowing the exact number of permutations helps determine "how likely" a specific call letter can occur.

Overall, these techniques form the foundation of numerous applications in mathematics and beyond, from fundamental data organization issues to complex probabilistic models.

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Most popular questions from this chapter

Odds are used in gambling games to make them fair. For example, if you rolled a die and won every time you rolled a \(6,\) then you would win on average once every 6 times. So that the game is fair, the odds of 5 to 1 are given. This means that if you bet \(\$ 1\) and won, you could win \(\$ 5 .\) On average, you would win \(\$ 5\) once in 6 rolls and lose \(\$ 1\) on the other 5 rolls - hence the term fair game. In most gambling games, the odds given are not fair. For example, if the odds of winning are really 20 to 1 the house might offer 15 to 1 in order to make a profit. Odds can be expressed as a fraction or as a ratio, such as \(\frac{5}{1}, 5: 1,\) or 5 to \(1 .\) Odds are computed in favor of the event or against the event. The formulas for odds are $$ \begin{array}{l} \text { Odds in favor }=\frac{P(E)}{1-P(E)} \\ \text { Odds against }=\frac{P(\bar{E})}{1-P(\bar{E})} \end{array} $$ In the die example, $$ \begin{aligned} &\text { Odds in favor of a } 6=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5} \text { or } 1: 5\\\ &\text { Odds against a } 6=\frac{\frac{5}{6}}{\frac{1}{6}}=\frac{5}{1} \text { or } 5: \end{aligned} $$ 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

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