Problem 21
Question
An old zip code in the United States consists of five digits. Find the total number of possible zip codes that: Have no repetitions.
Step-by-Step Solution
Verified Answer
\[P(10, 5) = 30{,}240\]
There are 30,240 possible zip codes with no repetitions.
1Step 1: Understand permutations
Permutations are used to count the number of ways to arrange a certain number of unique items. In this case, we have 10 unique digits (0-9) and 5 positions in the zip code.
2Step 2: Calculate the number of permutations without repetitions
To calculate the number of permutations without repetitions, we use the formula:
\(P(n, r) = \frac{n!}{(n-r)!}\)
where \(P(n, r)\) represents the number of permutations, \(n\) represents the total number of items to choose from, and \(r\) represents the number of items to be chosen.
In our case, \(n = 10\) since there are 10 unique digits, and \(r = 5\) because there are 5 positions in the zip code.
3Step 3: Substitute the values into the formula
Now substitute the values of \(n\) and \(r\) into the formula:
\(P(10, 5) = \frac{10!}{(10-5)!}\)
4Step 4: Calculate the factorials and solve
Calculate the factorials:
\(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3{,}628{,}800\)
\((10-5)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
Now divide \(10!\) by \(5!\):
\(P(10, 5) = \frac{3{,}628{,}800}{120}\)
5Step 5: Find the total number of possible zip codes
Finally, divide 3,628,800 by 120 to get the total number of possible zip codes without repetitions:
\(P(10, 5) = 30{,}240\)
So, there are a total of 30,240 possible zip codes with no repetitions.
Key Concepts
FactorialsCombinatoricsZip Code Problem
Factorials
Factorials are a fundamental concept in mathematics that are often used in permutations and combinations. A factorial, noted by an exclamation point, represents the product of all positive integers up to a given number. For example, the factorial of 5, written as 5!, is calculated as:
When using permutations like in our example, we often calculate factorials for both the total number and the difference between total items and chosen items, reflecting all possible combinations.
- 5! = 5 × 4 × 3 × 2 × 1 = 120
When using permutations like in our example, we often calculate factorials for both the total number and the difference between total items and chosen items, reflecting all possible combinations.
Combinatorics
Combinatorics is a branch of mathematics that studies counting, arrangement, and combination of objects. It is a powerful tool used to solve problems related to permutations and combinations. In the context of permutations, like our zip code problem, we are concerned with the different ways to arrange a set of items where the order matters.
To solve permutation problems, we use a specific formula:
To solve permutation problems, we use a specific formula:
- \[ P(n, r) = \frac{n!}{(n-r)!} \]
Zip Code Problem
The zip code problem explores a practical application of permutations and combinatorics. In this scenario, we wanted to calculate the number of unique five-digit zip codes possible using digits 0-9 without repeating any of the digits.
The question requires understanding how to apply permutation calculations because each zip code represents an order of digits. Given there are 10 distinct digits, we calculate how many ways we can select and arrange five of them.
This problem highlights the real-world applicability of permutations, helping us understand how mathematical theories define constraints and possibilities in everyday settings, such as creating zip codes.
The question requires understanding how to apply permutation calculations because each zip code represents an order of digits. Given there are 10 distinct digits, we calculate how many ways we can select and arrange five of them.
- The calculation involved is: \[ P(10, 5) = \frac{10!}{5!} \]
This problem highlights the real-world applicability of permutations, helping us understand how mathematical theories define constraints and possibilities in everyday settings, such as creating zip codes.
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Problem 21
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