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Chapter 4: Problem 24

During the Computer Daze special promotion, a customer purchasing a computer and printer is given a choice of three free software packages. There are 10 different software packages from which to select. How many different groups of software packages can be selected?

Short Answer

Expert verified
There are 120 different ways to select 3 software packages from 10.

Step by step solution

01

Understand the Problem

We need to determine how many different ways a customer can select 3 software packages from a total of 10 available packages. The order in which the packages are selected does not matter, so this is a combination problem.
02

Identify the Combination Formula

The combination formula is given by \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. In this problem, \( n = 10 \) and \( r = 3 \).
03

Plug Values Into the Formula

Substitute \( n = 10 \) and \( r = 3 \) into the combination formula: \( C(10, 3) = \frac{10!}{3! (10-3)!} = \frac{10!}{3! \times 7!} \).
04

Simplify the Factorials

Simplify the expression by calculating the factorials. We can write \( 10! = 10 \times 9 \times 8 \times 7! \) to cancel out \( 7! \) in the denominator: \( C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \).
05

Perform the Calculation

Calculate the expression \( \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \). First, calculate the numerator: \( 10 \times 9 = 90 \) and \( 90 \times 8 = 720 \). Then, calculate the denominator: \( 3 \times 2 = 6 \) and \( 6 \times 1 = 6 \). Therefore, \( \frac{720}{6} = 120 \).

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Key Concepts

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

Combination formula
In this problem, the goal is to find out in how many different ways you can select certain items from a larger set when the order doesn't matter. This is a classic example of a combination problem. To solve it, we use the combination formula. The combination formula is: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]Here's a brief explanation of what each symbol means:
  • The mathematical symbol \( C(n, r) \) represents the number of combinations possible when choosing \( r \) items from a total of \( n \) items.
  • The symbol "\(!\)" denotes a factorial.
  • \( n \) is the total number of available items.
  • \( r \) is the number of items to choose.
Using this formula helps determine the number of different groups you can form when picking \( r \) items from \( n \), without caring about which comes first.
Factorials
Factorials are a fundamental part of combinations and permutations. When we mention \( n! \), we're talking about the factorial of \( n \). It means you will multiply all whole numbers starting from \( n \) down to 1. For example:
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \( 3! = 3 \times 2 \times 1 = 6 \)
  • \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
Factorials help simplify calculations within the combination formula. They are especially useful since they allow us to cancel out terms, making the calculation easier to handle. For instance, in our previous example, with \( C(10, 3) \), knowing to express \( 10! \) as \( 10 \times 9 \times 8 \times 7! \) makes it easy to cancel \( 7! \) in the formula.
Combination problems
Combination problems are all about making selections. The focus is entirely on choosing, rather than arranging. Typical combination problems include situations where:
  • You are selecting a group of players for a team.
  • Picking cards from a deck but not caring about their order.
  • Choosing free software packages in a promotional offer, like in our given problem.
The key point in combination problems is the phrase "order doesn't matter." In contrast, permutation problems deal with arrangements where the sequence of choices is important. For combinations, just plug into the formula \( C(n, r) = \frac{n!}{r!(n-r)!} \), simplify any factorial needed, then calculate to find your answer. This approach makes it straightforward to determine how many possible ways exist to make your choices without worrying about their order.

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Most popular questions from this chapter

The qualified applicant pool for six management trainee positions consists of seven women and five men. (a) How many different groups of applicants can be selected for the positions? (b) How many different groups of trainees would consist entirely of women? (c) Probability extension: If the applicants are equally qualified and the trainee positions are selected by drawing the names at random so that all groups of six are equally likely, what is the probability that the trainee class will consist entirely of women?

(a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely? (b) Assign probabilities to the outcomes of the sample space of part (a). Do the probabilities add up to 1 ? Should they add up to \(1 ?\) Explain. (c) What is the probability of getting a number less than 5 on a single throw? (d) What is the probability of getting 5 or 6 on a single throw?

You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3\) on 1 st card and 10 on 2 nd ). (c) Find \(P(10\) on 1 st card and 3 on 2 nd ). (d) Find the probability of drawing a 10 and a 3 in either order.

There are three nursing positions to be filled at Lilly Hospital. Position 1 is the day nursing supervisor; position 2 is the night nursing supervisor; and position 3 is the nursing coordinator position. There are 15 candidates qualified for all three of the positions. Determine the number of different ways the positions can be filled by these applicants.

In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: $$\begin{array}{lllr} \hline & \text { Sale } & \text { No Sale } & \text { Row Total } \\ \hline \text { Aggressive } & 270 & 310 & 580 \\ \text { Passive } & 416 & 164 & 580 \\ \text { Column Total } & 686 & 474 & 1160 \\ \hline \end{array}$$ Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: \(A=\) aggressive approach, \(P a=\) passive approach, \(S=\) sale, \(N=\) no sale. So, \(P(A)\) is the probability that an aggressive approach was used, and so on. (a) Compute \(P(S), P(S \mid A)\), and \(P(S \mid P a)\). (b) Are the events \(S=\) sale and \(P a=\) passive approach independent? Explain. (c) Compute \(P(A\) and \(S)\) and \(P(P a\) and \(S)\). (d) Compute \(P(N)\) and \(P(N \mid A)\). (e) Are the events \(N=\) no sale and \(A=\) aggressive approach independent? Explain. (f) Compute \(P(A\) or \(S)\).
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