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A pollster randomly selected 4 of 10 available people. How many different groups of 4 are possible?

Short Answer

Expert verified
210 groups

Step by step solution

01

Understanding the Problem

We need to find how many different ways 4 people can be chosen from a group of 10. Since the order does not matter, we are looking for the number of combinations.
02

Identify the Combinatorial Formula

For a problem like this, we use the combination formula. The number of ways to choose \( r \) elements from a set of \( n \) elements is given by: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] In this problem, \( n = 10 \) and \( r = 4 \).
03

Calculate the Factorials

Calculate the factorials required for the formula: - \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)- \( 4! = 4 \times 3 \times 2 \times 1 \)- \( (10-4)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
04

Apply the Formula

Substitute the calculated factorials into the combination formula: \[ C(10, 4) = \frac{10!}{4! \, 6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4 \times 3 \times 2 \times 1 \times 6!} \]Notice that \( 6! \) is on both the numerator and the denominator, so they cancel each other out.
05

Simplify and Calculate

Now simplify the remaining expression: \[ \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \]Calculate the numerator: - \( 10 \times 9 = 90 \)- \( 90 \times 8 = 720 \)- \( 720 \times 7 = 5040 \)Calculate the denominator: - \( 4 \times 3 = 12 \)- \( 12 \times 2 = 24 \)- \( 24 \times 1 = 24 \)Finally, divide the numerator by the denominator: \[ \frac{5040}{24} = 210 \]
06

Conclusion

Thus, there are 210 different combinations or groups of 4 people that can be made from the group of 10.

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Key Concepts

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

Combinations
In the world of combinatorics, one of the most fundamental concepts is that of combinations. Combinations are all about selecting items from a larger group, where the order in which you select them doesn't matter. For example, in our exercise, we are interested in how many different ways we can pick 4 people out of a group of 10. Once you decide which 4 people to select, it doesn't matter who was picked first or last; the group is the same.

To find combinations, we use the combination formula, denoted as \( C(n, r) \). This represents the number of ways to choose \( r \) elements from a set of \( n \) elements. The formula is:
  • \[ C(n, r) = \frac{n!}{r!(n-r)!} \]
In the context of our exercise, \( n = 10 \) and \( r = 4 \). Plugging in these numbers gives us the number of ways to create different groups. The concept of combinations is essential for solving problems where the arrangement of selected items isn't important, unlike in permutations.
Factorials
Factorials are crucial when calculating combinations. A factorial, denoted by the symbol \(!\), is the product of all positive integers up to a specified number. For example, \(10!\) means multiplying all numbers from 10 down to 1. It's calculated as follows:
  • \[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]
Factorials grow very rapidly, even with small increases in the number. They are used in many formulas due to their properties of counting the total possible arrangements of a set. In the combination formula, factorials in the denominator (like \(r!\) and \((n-r)!\)) help remove the effect of order, leaving us with just the number of unique selections.

In the exercise, we needed \(10!\), \(4!\), and \(6!\) since \(n = 10\) and \(r = 4\); calculating these factorials was a key part of using the combination formula.
Probability
While this particular exercise focuses on combinatorics, it's essential to understand how combinations connect to probability. Probability is all about the likelihood of an event occurring. When dealing with combinations, probability allows us to determine how likely a specific grouping or selection is to happen.

To calculate probability using combinations, you often need to find the number of successful outcomes (using combinations) divided by the total number of possible outcomes. For instance, if you need to calculate the probability of selecting a certain group of people from a larger set in our exercise concept, you would first figure out all possible combinations. Then, you'd identify which of these combinations meet your criteria and compare that to the total number of combinations.
  • This serves as a bridge between simple counting and assessing chances, blending combinatorics with statistics.
Understanding how probability relates to combinations allows you to predict and assess real-world scenarios where outcomes may seem random or unpredictable at first glance.

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Most popular questions from this chapter

An Internet company located in Southern California has season tickets to the Los Angeles Lakers basketball games. The company president always invites one of the four vice presidents to attend games with him, and claims he selects the person to attend at random. One of the four vice presidents has not been invited to attend any of the last five Lakers home qames. What is the likelihood this could be due to chance?

A sample of 40 oil industry executives was selected to test a questionnaire. One question about environmental issues required a yes or no answer. a. What is the experiment? b. List one possible event. c. Ten of the 40 executives responded yes. Based on these sample responses, what is the probability that an oil industry executive will respond yes? d. What concept of probability does this illustrate? e. Are each of the possible outcomes equally likely and mutually exclusive?

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There are 20 families living in the Willbrook Farms Development. Of these families, 10 prepared their own federal income taxes for last year, 7 had their taxes prepared by a local professional, and the remaining 3 were done by H\&R Block. a. What is the probability of selecting a family that prepared their own taxes? b. What is the probability of selecting two families, both of which prepared their own taxes? c. What is the probability of selecting three families, all of which prepared their own taxes? d. What is the probability of selecting two families, neither of which had their taxes prepared by H\&R Block?

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