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Chapter 9: Problem 26

A combination lock will open when the right choice of three numbers (from \( 1 \) to \( 50 \), inclusive) is selected. How many different lock combinations are possible?

Short Answer

Expert verified
The total number of different lock combinations is \( 50^3 = 125000 \).

Step by step solution

01

Understanding the problem

Consider each number as a distinct choice and presume repetition is allowed. The first number can be any of 50 choices, and so can be the second and third.
02

Calculation

Calculate the total choices by multiplying the number of choices for each number. As there are 50 choices for the first number, 50 for the second and 50 for the third, the total number of permutations can be represented as \( 50*50*50 \).
03

Final answer

Executing the multiplication (50*50*50), we can find the total number of different combinations.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arranging, and combining objects. In the context of combination locks, combinatorics helps determine how many possible ways you can arrange a set of numbers.
It's like playing with different arrangements and selections of numbers.
  • Combinatorics deals with the study of finite or countable discreet systems, which means systems that can eventually be counted one by one.
  • It examines how objects can be selected and arranged in different combinations, either by themselves or in groups.
  • Its applications stretch from simple tasks like figuring out lock codes to complex tasks in computer science and cryptography.
Understanding the basics of combinatorics will help you solve problems involving counts and arrangements, like determining the number of combinations possible for a lock.
Repetition Allowed
The concept of repetition allowed means you can use the same number more than once in a combination. In our lock example, this means any of the numbers from 1 to 50 can appear multiple times in the combination.
For instance, "5-5-5" is a valid combination.
  • This concept hugely increases the number of possible combinations.
  • If repetition was not allowed, each of the numbers would have to be different, drastically reducing the possibilities.
  • With repetition permitted, each position in a choice can be filled without restriction, allowing greater flexibility and more options.
Embracing repetition allows you to explore a wider array of combinations, crucial for tasks like setting locks or coding with finite sets of numbers.
Permutations
Permutations refer to the arrangement of elements in a particular order. When we consider permutations in combination locks with repetition, each position in the sequence (like our three-number lock) can have any number from 1 to 50.
It's about considering the order of each choice.
  • A permutation considers the importance of sequence, unlike a simple combination where order doesn't matter.
  • For our lock, a sequence like "1-2-3" is different from "3-2-1" because the order of numbers affects the result.
  • The number of permutations available is calculated by multiplying the number of options per position, which in this case, was found as 50 choices raised to the power of the number of positions (three), so the formula for calculation was \( 50^3 \).
Understanding permutations help you appreciate why order is significant and how arranging elements affects outcomes in tasks involving ordered selections, like a combination lock.

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Most popular questions from this chapter

In Exercises 9 - 14, determine the sample space for the experiment. A taste tester has to rank three varieties of yogurt, \( A \), \( B \),and \( C \), according to preference.

American roulette is a game in which a wheel turns on a spindle and is divided into \( 38 \) pockets.Thirty-six of the pockets are numbered \( 1-36 \), of which half are red and half are black. Two of the pockets are green and are numbered \( 0 \) and \( 00 \) (see figure). The dealer spins the wheel and a small ball in opposite directions. As the ball slows to a stop, it has an equal probability of landing in any of the numbered pockets. (a) Find the probability of landing in the number \( 00 \) pocket. (b) Find the probability of landing in a red pocket. (c) Find the probability of landing in a green pocket or a black pocket. (d) Find the probability of landing in the number \( 14 \) pocket on two consecutive spins. (e) Find the probability of landing in a red pocket on three consecutive spins.

Without calculating the numbers, determine which of the following is greater. Explain. (a) The number of combinations of \( 10 \) elements taken six at a time (b) The number of permutations of \( 10 \) elements taken six at a time

Consider a group of people. (a) Explain why the following pattern gives the probabilities that the people have distinct birthdays. \( n = 2: \dfrac{365}{365} \cdot \dfrac{364}{365} = \dfrac{365 \cdot 364}{365^2} \) \( n = 3: \dfrac{365}{365} \cdot \dfrac{364}{365} \cdot \dfrac{363}{365} = \dfrac{365 \cdot 364 \cdot 363}{365^3} \) (b) Use the pattern in part (a) to write an expression for the probability that \( n = 4 \) people have distinct birthdays. (c) Let \( P_n \) be the probability that the \( n \) people have distinct birthdays. Verify that this probability can be obtained recursively by \( P_1 = 1 \) and \( P_n = \dfrac{365 - (n - 1)}{365} P_{n - 1} \). (d) Explain why \( Q_n = 1 - P_n \) gives the probability that at least two people in a group of \( n \) people have the same birthday. (e) Use the results of parts (c) and (d) to complete the table. (f) How many people must be in a group so that the probability of at least two of them having the same birthday is greater than \( \dfrac{1}{2} \)? Explain.

In Exercises 91 - 94, prove the identity. \( _nC_n = _nC_0 \)
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