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How many ways can an adviser choose 4 students from a class of 12 if they are all assigned the same task? How many ways can the students be chosen if they are each given a different task?

Short Answer

Expert verified
495 ways for the same task; 11880 ways for different tasks.

Step by step solution

01

Understanding the Problem

We need to determine the number of ways to choose students in two scenarios: 1) All students are given the same task, which involves choosing combinations; 2) Students are given different tasks, which involves permutations.
02

Choosing 4 Students for the Same Task

Since the task is the same for each student, we use combinations to select the students. The formula for combinations is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \(n\) is the total number of students, and \(k\) is the number of students to choose. Here, we calculate \( \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = 495\).
03

Assigning Different Tasks to 4 Students

Since each student gets a different task, we use permutations. After choosing 4 students out of 12, we need to assign the tasks, which is calculated using \(4!\). First, compute \( \binom{12}{4} = 495\) as before, then multiply by \(4! = 24\) for the task assignments. Therefore, the total number of ways is \( 495 \times 24 = 11880 \).

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Key Concepts

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

Combinations
When it comes to selecting a group of items where the order doesn't matter, we use combinations. In the scenario where an adviser chooses 4 students from a class of 12 to perform the same task, we're dealing with combinations.The formula for combinations is given by:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where:
  • \(n\) is the total number of items to choose from (in our example, 12 students)
  • \(k\) is the number of items to choose (in our example, 4 students)
Using this formula, the number of ways to choose 4 students out of 12 is:\[ \binom{12}{4} = \frac{12!}{4! \times 8!} = 495 \]Here, the factorial operation \(!\) denotes the product of all positive integers up to the given number, making the calculation feasible and systematic for any similar problem.
Permutations
Permutations are all about order. When students are assigned different tasks, the order in which they are assigned matters, and thus permutations come into play.In a scenario where each selected student from the group receives a unique task, the formula stretches to accommodate order:After choosing 4 students from 12, these students can be assigned tasks in permutations of themselves.The general formula for permutations when selecting \(k\) items from \(n\) is:\[ P(n, k) = \frac{n!}{(n-k)!} \]However, in our exercise, once the 4 students (from the combination \(\binom{12}{4} = 495\)) are chosen, each student can be assigned one of the 4 unique tasks. This is calculated simply as:\[ 4! = 24 \]So, the number of ways is:\[ \binom{12}{4} \times 4! = 495 \times 24 = 11880 \]The combination of selecting the group and then permuting their order makes permutation greatly powerful whenever specifics of order count.
Factorial
The term 'factorial' is key to comprehending permutations and combinations comprehensively. Factorial of a non-negative integer \(n\), denoted as \(n!\), represents the product of all positive integers up to \(n\).For instance:
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
  • \(3! = 3 \times 2 \times 1 = 6\)
  • Factorial of zero is 1, \(0! = 1\)
Factorial forms the backbone of computations involving permutations and combinations.In our exercise, it calculates the number of possible arrangements of a selected group or the number of ways to choose a smaller group from a large set. The factorial grows rapidly with larger numbers, making it essential for determining possibilities whenever grouping or ordering is in play.

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Most popular questions from this chapter

The frequency distribution shown here illustrates the number of medical tests conducted on 30 randomly selected emergency patients. $$ \begin{array}{lc} \begin{array}{c} \text { Number of tests } \\ \text { performed } \end{array} & \begin{array}{c} \text { Number of } \\ \text { patients } \end{array} \\ \hline 0 & 12 \\ 1 & 8 \\ 2 & 2 \\ 3 & 3 \\ 4 \text { or more } & 5 \end{array} $$ If a patient is selected at random, find these probabilities. a. The patient has had exactly 2 tests done. b. The patient has had at least 2 tests done. c. The patient has had at most 3 tests done. d. The patient has had 3 or fewer tests done. e. The patient has had 1 or 2 tests done.

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First-year students at a particular college must take one English class, one class in mathematics, a first-year seminar, and an elective. There are 2 English classes to choose from, 3 mathematics classes, 5 electives, and everyone takes the same first-year seminar. Represent the possible schedules, using a tree diagram.

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Two bags contain marbles. Bag 1 contains 1 black marble and 9 white marbles. Bag 2 contains 1 black marble and \(x\) white marbles. If you choose a bag at random, then choose a marble at random, the probability of getting a black marble is \(\frac{2}{15}\). How many white marbles are in bag \(2 ?\)

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