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How many ways are there to select 3 candidates from 8 equally qualified recent graduates for openings in an accounting firm?

Short Answer

Expert verified
There are 56 different ways to select 3 candidates from 8 equally qualified recent graduates.

Step by step solution

01

Understanding Combinations

A combination is a selection of items where order does not matter. Mathematically, a combination can be calculated using the formula \(\binom{n}{r} = \frac{n!}{(n-r)!r!}\) , where \(n\) is the total number of items, \(r\) is the number of items to select, and \(!\) denotes factorial, which is the product of an integer and all the integers below it (e.g.\(5! = 5*4*3*2*1 = 120\)).
02

Calculating Factorials

You need to calculate the values of \(8!\), \((8-3)!\), and \(3!\) to use in the combination formula. Calculating these you get: \(8! = 40320\), \((8-3)! = 5! = 120\), and \(3! = 6\).
03

Applying the Combination Formula

Putting the values you calculated in Step 2 into the combination formula you get: \(\binom{8}{3} = \frac{8!}{5!*3!} = \frac{40320}{120*6} = \frac{40320}{720} = 56\).

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Key Concepts

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

Factorial Calculations
A foundational concept in the realm of combinations and probability is understanding factorial calculations. A factorial, denoted by an exclamation point (!), represents the product of an integer and all the positive integers below it. For example, when we want to calculate the value of 5 factorial, denoted as \(5!\), we multiply 5 by every integer less than itself down to 1:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120. \]
Factorials grow extremely quickly with larger numbers. They are integral in computing combinations because they account for the total number of ways to arrange a set number of items. Recognizing how to do factorial calculations is vital as it is frequently the base for problems in permutations and combinations within combinatorics. A simple way to calculate small factorials is to write down the sequence but for larger numbers, it's often easier to use a calculator or factorial function in programming languages.
Combination Formula
In probability, the combination formula is pivotal for finding the number of ways to select items when the order doesn't matter. The combination formula is:
\[ \binom{n}{r} = \frac{n!}{(n-r)!r!} \]
Where \(n\) represents the total number of items and \(r\) is the number of items you want to select. Let's break it down:
  • \(n!\) gives us all possible arrangements (permutations) of \(n\) items.
  • \((n-r)!\) then adjusts for the permutations of the items not chosen, reducing the total count.
  • Finally, \(r!\) accounts for the order of selection not mattering by dividing out the permutations of the selected items.
By putting these pieces together, the formula calculates the number of unique sets you can form. It's a brilliant display of simplifying the process of selection, subtracting out irrelevant permutations, and leaving us with just combinations.
Combinatorics
Combinatorics is a field of mathematics primarily concerned with counting, both as a means and an end in obtaining results, as well as certain properties of finite structures. It's often considered a part of discrete mathematics and involves studying combinations, permutations, and configurations.
In our example involving recent graduates, combinatorics guides us through various ways we can select 3 out of 8 candidates. This area of math doesn't only deal with numbers; it is also interested in the arrangement, selection, and combination of items within a set. By mastering combinatorics, you learn to approach problems in a structured manner, easily breaking down complex scenarios into manageable calculations.
One of the real beauties of combinatorics is its wide applicability, from creating computer algorithms to predicting probabilities in genetics. It teaches us to navigate through an array of possibilities, emphasizes logical reasoning, and develops our ability to analyze and solve various types of problems.

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Most popular questions from this chapter

If \(S=\\{0,1,2,3,4,5,6,7,8,9\\}\) and \(A=\) \(\\{0,2,4,6,8\\}, B=\\{1,3,5,7,9\\}, C=\\{2,3,4,5\\},\) and \(D=\\{1,6,7\\},\) list the elements of the sets corresponding to the following events: (a) \(A \mathrm{U} C\); (b) \(A \cap B\) (c) \(C^{\prime}\); (d) \(\left(C^{\prime} \cap D\right) \cup B\); (e) \((S \cap C)^{\prime}\) (f) \(A \cap C\) n \(D^{\prime}\).

A truth serum has the property that \(90 \%\) of the guilty suspects are properly judged while, of course, \(10 \%\) of guilty suspects are improperly found innocent. On the other hand, innocent suspects are misjudged \(1 \%\) of the time. If the suspect was selected from a group of suspects of which only \(5 \%\) have ever committed a crime, and the serum indicates that he is guilty, what is the probability that he is innocent?

A California study concluded that by following 7 simple health rules a man's life can be extended by 11 years on the average and a woman's life by 7 years. These 7 rules are as follows: no smoking, regular exercise, use alcohol moderately, get, 7 to 8 hours of sleep, maintain proper weight, eat. breakfast, and do not eat between meals. In how many ways can a person adopt five of these rules to follow (a) If the person presently violates all 7 rules? (b) If the person never drinks and always eats breakfast?

In a fuel economy study, each of 3 race cars is tested using 5 different brands of gasoline at 7 test sites located in different regions of the country. If 2 drivers are used in the study, and test runs are made once under each distinct set of conditions, how many test runs are needed?

Nine people are going on a skiing trip in 3 cars that hold \(2,4,\) and 5 passengers, respectively. In how many ways is it possible to transport the 9 people to the ski lodge, using all cars?

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