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A company is creating three new divisions and seven managers are eligible to be appointed head of a division. How many different ways could the three new heads be appointed?

Short Answer

Expert verified
There are 35 different ways to appoint the three new heads.

Step by step solution

01

Determine Combination Formula

To find out how many ways the three heads can be selected out of seven managers, we need to use the combination formula. The combination formula is used when the order of selection does not matter, and it is given by: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
02

Identify Values for the Formula

In this problem, we are selecting 3 managers out of 7 to be the heads. So, \( n = 7 \) and \( r = 3 \). Substitute these values into the combination formula.
03

Compute Factorials

Calculate the factorials: \( 7! \), \( 3! \), and \( (7-3)! \). 1. \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \) 2. \( 3! = 3 \times 2 \times 1 = 6 \) 3. \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
04

Plug Values into Combination Formula

Substitute the factorial values into the combination formula: \[ \binom{7}{3} = \frac{7!}{3! \times 4!} = \frac{5040}{6 \times 24} \]
05

Simplify the Expression

First, calculate the denominator: - Multiply: \( 6 \times 24 = 144 \)- Divide: \( 5040 \div 144 = 35 \)Thus, \[ \binom{7}{3} = 35 \]
06

Interpret Result

The calculation tells us there are 35 different ways to appoint the three heads of the divisions out of seven eligible managers.

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Key Concepts

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

Factorials
Understanding the concept of factorials is crucial when solving combination problems. A factorial, denoted by the symbol "!", is the product of all positive integers up to a given number. For instance, the factorial of 4, written as 4!, is calculated by multiplying 4 by every positive integer below it:
  • 4! = 4 × 3 × 2 × 1 = 24
Factorials serve as the building blocks of the combination formula. They allow us to calculate the number of different ways in which a certain number of items can be arranged. The larger the number, the more complex the factorial, but they always follow this simple multiplication rule.
Factorials can quickly become large numbers. Thankfully, many calculators have a factorial function that can compute them instantly. In problems where combinations or permutations are considered, understanding how to break them down into their factorial components can simplify the solution process.
Combination Formula
The combination formula is a key concept in probability and statistics, providing a way to determine how many ways a set number of items can be selected from a larger group without regard to order. The combination formula is written as:
  • \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
Here, \( n \) represents the total number of items, and \( r \) is the number of items to be chosen. The expression \( \binom{n}{r} \) is read as "n choose r," denoting the number of combinations possible.
In our original exercise, we wanted to determine how many different ways three managers out of seven could be appointed. Using the values \( n = 7 \) and \( r = 3 \), the combination formula helps compute this efficiently by including factorial calculations for \( n \), \( r \), and \( n-r \).
This formula is particularly useful when the selection order doesn't matter—differentiating it from permutations. Practicing the use of this formula in various scenarios solidifies understanding and helps in recognizing pattern-based math problems.
Permutations vs Combinations
Permutations and combinations are two methods to compute how objects can be selected. The key difference lies in whether the order of selection matters.
  • Permutations involve scenarios where order is important. If you were picking three medals—gold, silver, and bronze—from a group, which medal goes to which person matters, so permutations are used. The formula for permutations of \( n \) items taken \( r \) at a time is:\[ P(n, r) = \frac{n!}{(n-r)!} \]
  • Combinations, on the other hand, are used when the order doesn't matter. A simple example would be forming a committee where it doesn't matter who is chosen first or last. The combination formula, \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), assumes all selections are equally valid regardless of order.
To decide between using permutations or combinations, ask whether the sequence of picking is crucial. If not, and only a selection is needed, like in our exercise appointing division heads, combinations are appropriate. Understanding these differences aids in solving various statistical, probabilistic, and day-to-day problems more effectively.

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Most popular questions from this chapter

Joe Mauer of the Minnesota Twins had the highest batting average in the 2006 Major League Baseball season. His average was . \(347 .\) So assume the probability of getting a hit is . 347 for each time he batted. In a particular game assume he batted three times. a. This is an example of what type of probability? b. What is the probability of getting three hits in a particular game? c. What is the probability of not getting any hits in a game? d. What is the probability of getting at least one hit?

The probabilities of the events \(A\) and \(B\) are .20 and .30 , respectively. The probability that both \(A\) and \(B\) occur is .15. What is the probability of either \(A\) or \(B\) occurring?

The state of Maryland has license plates with three numbers followed by three letters. How many different license plates are possible?

In each of the following cases, indicate whether classical, empirical, or subjective probability is used. a. A baseball player gets a hit in 30 out of 100 times at bat. The probability is .3 that he gets a hit in his next at bat. b. A seven-member committee of students is formed to study environmental issues. What is the likelihood that any one of the seven is chosen as the spokesperson? c. You purchase one of 5 million tickets sold for Lotto Canada. What is the likelihood you win the \(\$ 1\) million jackpot? d. The probability of an earthquake in northern California in the next 10 years is . 80 .

Refer to Exercise \(56,\) but assume there are 10 homes in the Quail Creek area and four of them have a security system. Three homes are selected at random: a. What is the probability all three of the selected homes have a security system? b. What is the probability none of the three selected homes have a security system? c. What is the probability at least one of the selected homes has a security system? d. Did you assume the events to be dependent or independent?

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