/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Evaluate each expression. \(\f... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each expression. \(\frac{_{5} C_{1} \cdot_{7} C_{2}}{_{12} C_{3}}\)

Short Answer

Expert verified
The evaluation of the originally given expression, \(\frac{_5C_1 * _7C_2}{_12C_3}\), is approximately 0.47727, rounded to the nearest hundred-thousandth.

Step by step solution

01

Calculate first combination

To calculate the first combination, _5C_1, plug the numbers into the formula for combinations, yielding \[\frac{5!}{1!(5-1)!} = \frac{5!}{1*4!}\]. The factorial \(5!\) is \(5*4*3*2*1\), and \(4!\) is \(4*3*2*1.\) Thus, \(\frac{5!}{1*4!}\) simplifies to 5.
02

Calculate second combination

To calculate the second combination, _7C_2, plug numbers into the formula for combinations, yielding \[\frac{7!}{2!(7-2)!} = \frac{7!}{2*5!}\]. The factorial \(7!\) is \(7 * 6 * 5 * 4 * 3 * 2 * 1\), and \(5!\) is \(5 * 4 * 3 * 2 * 1\). So, \(\frac{7!}{2*5!}\) simplifies to \(\frac{7 * 6}{2} = 21.\)
03

Calculate third combination

To calculate the third combination, _12C_3, plug the numbers into the combination formula, yielding \[\frac{12!}{3!(12-3)!} = \frac{12!}{3*9!}\]. The factorial \(12!\) is \(12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1\), and \(9!\) is \(9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1\). So, \(\frac{12!}{3*9!}\) simplifies to \(\frac{12 * 11 * 10}{3 * 2 * 1} = 220.\)
04

Perform final calculation

Now, with the combinations calculated, we just need to plug them into the original problem statement and do the division. So this turns into \(\frac{5 * 21}{220}\), which simplifies to \(\frac{105}{220}\), and further simplifies to 0.47727 when carried out to five decimal places. The solution is approximated to the nearest hundred-thousandth for more precise results.

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Key Concepts

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

Permutations
Permutations are one of the fundamental concepts in combinatorics. They refer to the different ways in which a set of items can be ordered or arranged. Importantly, permutations distinguish between various orderings, making this concept essential when the sequence of items is crucial.
Imagine you have a small collection of objects like letters or numbers. The possible permutations of these items are dependent not just on the count of the items but also on the order in which they appear.
  • A permutation of 3 letters 'A', 'B', 'C' in all possible orders is: ABC, ACB, BAC, BCA, CAB, CBA.
  • Each different arrangement counts as a unique permutation.
The mathematical expression for permutations can be calculated using factorials, specifically represented as \(P(n, r)\) for arranging \(r\) objects out of \(n\) available ones.
This is mathematically defined as \( \frac{n!}{(n-r)!} \). Permutations are very useful in scenarios like seating arrangements and order-sensitive tasks.
Factorials
Factorials are a mathematical operation that forms a core part of many combinatorial calculations, including permutations and combinations. Written as \(n!\), a factorial is the product of all positive integers from \(1\) up to \(n\).
For instance:
  • \(3! = 3 \times 2 \times 1 = 6\)
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
  • \(0! = 1\), by definition
Factorials rapidly increase in value with larger numbers, thus influencing combinatorial calculations significantly. Their use in permutations and combinations helps determine the total number of possible outcomes or arrangements efficiently. Knowing how to compute and simplify factorials is pivotal in tackling problems involving large collections of items, as they aid in reducing complex calculations to simpler products.
Binomial Coefficient
The binomial coefficient, symbolized as \( \binom{n}{r} \), is a key concept in combinatorics that gives the number of ways \(r\) items can be chosen from \(n\) items, without considering the order. This is also known as a combination, contrasting with permutations, as here the sequence of items doesn't matter.
The mathematical formula used to calculate a binomial coefficient is \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \].
Here are some key properties:
  • The binomial coefficient tells us about possible group selections.
  • The calculation involves dividing the total arrangements by the permutations within selected groups.
For example, choosing 2 team members out of 5 without orientation concerns, it's calculated as \( \binom{5}{2} = \frac{5!}{2!3!} = 10\).
Understanding binomial coefficients is crucial in probability, statistical calculations, and various combinatorial constructs, as it simplifies accounting for group formations from larger pools.

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Most popular questions from this chapter

Explaining the Concepts The president of a large company with \(10,000\) employees is considering mandatory cocaine testing for every employee. The test that would be used is \(90 \%\) accurate, meaning that it will detect \(90 \%\) of the cocaine users who are tested, and that \(90 \%\) of the nonusers will test negative. This also means that the test gives \(10 \%\) false positive. Suppose that \(1 \%\) of the employees actually use cocaine. Find the probability that someone who tests positive for cocaine use is, indeed, a user. Hint: Find the following probability fraction: the number of employees who test positive and are cocaine users the number of employees who test positive This fraction is given by $$ 90 \% \text { of } 1 \% \text { of } 10,000 $$ the number who test positive who actually use cocaine plus the number who test positive who do not use cocaine What does this probability indicate in terms of the percentage of employees who test positive who are not actually users? Discuss these numbers in terms of the issue of mandatory drug testing. Write a paper either in favor of or against mandatory drug testing, incorporating the actual percentage accuracy for such tests.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Beginning at 6: 45 A.M., a bus stops on my block every 23 minutes, so I used the formula for the \(n\) th term of an arithmetic sequence to describe the stopping time for the \(n\) th bus of the day.

Use the formula for the value of an annuity to solve Exercises 77–84. Round answers to the nearest dollar. To save money for a sabbatical to earn a master's degree, you deposit \(\$ 2000\) at the end of each year in an annuity that pays \(7.5 \%\) compounded annually. a. How much will you have saved at the end of five years? b. Find the interest.

In Exercises \(99-100,\) use a graphing utility to graph the function. Determine the horizontal asymptote for the graph of fand discuss its relationship to the sum of the given series. Function Series $$ f(x)=\frac{2\left[1-\left(\frac{1}{3}\right)^{x}\right]}{1-\frac{1}{3}} \quad 2+2\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)^{2}+2\left(\frac{1}{3}\right)^{3}+\cdots $$

Here are two ways of investing \(\$ 40,000\) for 25 years. \(\begin{array}{cccc}{\text { Lump-Sum Deposit }} & {\text { Rate }} & {\text { Time }} \\ {\$ 40,000} & {6.5 \% \text { compounded }} & {25 \text { years }} \\\ {} & {\text { annually }}\end{array}\) $$ \begin{array}{ll} {\text { Periodic Deposits }} & {\text { Rate } \quad \text { Time }} \\ {\$ 1600 \text { at the end }} & {6.5 \% \text { compounded } 25 \text { years }} \\ {\text { of each year }} & {\text { annually }} \end{array} $$ After 25 years, how much more will you have from the lump-sum investment than from the annuity?

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