/*! 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 18 Compute \(\mathrm{C}_{8,3} .\)... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 18

Compute \(\mathrm{C}_{8,3} .\)

Short Answer

Expert verified
\( \mathrm{C}_{8,3} = 56 \)

Step by step solution

01

Understanding the Problem

We need to compute the combination, which is denoted as \( \mathrm{C}_{8,3} \). This means we are looking for the number of ways to choose 3 items from a set of 8 items without regard to the order.
02

Using the Combination Formula

The formula to calculate a combination is given by: \[ \mathrm{C}_{n,k} = \frac{n!}{k! \, (n-k)!} \] where \( n \) is the total number of items (8 in this case), and \( k \) is the number of items to choose (3 in this case). So, we use \[ \mathrm{C}_{8,3} = \frac{8!}{3! \, (8-3)!} \]
03

Calculating Factorials

First, compute the factorials required for the formula:- \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)- \( 3! = 3 \times 2 \times 1 = 6 \)- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
04

Plugging Values into the Formula

Substitute these values back into the combination formula: \[ \mathrm{C}_{8,3} = \frac{8!}{3! \, 5!} = \frac{40320}{6 \times 120} \]
05

Performing the Division

Calculate the denominator: \[ 6 \times 120 = 720 \]Now, divide the numerator by the denominator: \[ \mathrm{C}_{8,3} = \frac{40320}{720} = 56 \]

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Key Concepts

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

Understand the Combination Formula
The combination formula is essential in combinatorics, especially when you want to find out how many ways you can choose a certain number of items from a larger set without considering the order. This is perfect for scenarios where the order doesn't matter, like choosing toppings for a pizza or forming a committee from a group of people. The formula is expressed as:
  • \( \mathrm{C}_{n,k} = \frac{n!}{k! \times (n-k)!} \)
Here:
  • \( n \) is the total number of items you have to choose from.
  • \( k \) is the number of items you want to choose.
Understanding this formula will help you solve a wide variety of combination problems in mathematics and real-life applications.
It highlights that it's the selection, not the arrangement, that's crucial.
Exploring Factorials
Factorials play a vital role in calculating combinations. The factorial of a non-negative integer \( n \), written as \( n! \), is the product of all positive integers less than or equal to \( n \). For example,
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \( 3! = 3 \times 2 \times 1 = 6 \)
The concept of factorials is built into the combination formula to handle the different permutations of selecting items and to ensure that the calculation yields the right number of combinations. Factorials grow very quickly, which is why they are so effective in these types of problems.
Knowing how to calculate these efficiently is key when you work with larger numbers.
Mastering Calculating Combinations
To calculate combinations, comprehension of both the combination formula and factorials is necessary. Once you grasp these concepts, applying them is straightforward. For example, let's compute \( \mathrm{C}_{8,3} \).First, identify your \( n \) and \( k \):
  • \( n = 8 \)
  • \( k = 3 \)
Insert these into the formula:
  • \( \mathrm{C}_{8,3} = \frac{8!}{3! \times (8-3)!} = \frac{8!}{3! \times 5!} \)
Next, find each factorial:
  • \( 8! = 40320 \)
  • \( 3! = 6 \)
  • \( 5! = 120 \)
With these numbers, plug them back into the formula:
  • \( \mathrm{C}_{8,3} = \frac{40320}{6 \times 120} \)
Calculate the denominator \( 6 \times 120 = 720 \). Finally, divide to find:
  • \( \mathrm{C}_{8,3} = \frac{40320}{720} = 56 \)
Through these steps, you've successfully found there are 56 ways to choose 3 items from a set of 8 items.
By practicing this process, the task of computing combinations becomes more intuitive and less daunting.

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Most popular questions from this chapter

At Litchfield College of Nursing, \(85 \%\) of incoming freshmen nursing students are female and \(15 \%\) are male. Recent records indicate that \(70 \%\) of the entering female students will graduate with a BSN degree, while \(90 \%\) of the male students will obtain a BSN degree. If an incoming freshman nursing student is selected at random, find (a) \(P\) (student will graduate I student is female). (b) \(P(\) student will graduate and student is female). (c) \(P\) (student will graduate \(\mid\) student is male). (d) \(P\) (student will graduate and student is male). (e) \(P\) (student will graduate). Note that those who will graduate are either males who will graduate or females who will graduate. (f) The events described by the phrases "will graduate and is female" and "will graduate, given female" seem to be describing the same students. Why are the probabilities \(P\) (will graduate and is female) and \(P\) (will graduate female) different?

Expand Your Knowledge: Odds in Favor Sometimes probability statements are expressed in terms of odds. The odds in favour of an event \(A\) are the ratio \(\frac{P(A)}{P(\text { not } A)}=\frac{P(A)}{P\left(A^{c}\right)}\). For instance, if \(P(A)=0.60\), then \(P\left(A^{c}\right)=0.40\) and the odds in favor of \(A\) are \(\frac{0.60}{0.40}=\frac{6}{4}=\frac{3}{2}\), written as 3 to 2 or \(3: 2\) (a) Show that if we are given the odds in favor of event \(A\) as \(n: m\), the probability of event \(A\) is given by \(P(A)=\frac{n}{n+m} .\) Hint: Solve the equation \(\frac{n}{m}=\frac{P(A)}{1-P(A)}\) for \(P(A) .\)

Can you wiggle your ears? Use the students in your statistics class (or a group of friends) to estimate the percentage of people who can wiggle their ears. How can your result be thought of as an estimate for the probability that a person chosen at random can wiggle his or her ears? Comment: National statistics indicate that about \(13 \%\) of Americans can wiggle their ears (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York).

A recent Harris Poll survey of 1010 U.S. adults selected at random showed that 627 consider the occupation of firefighter to have very great prestige. Estimate the probability (to the nearest hundredth) that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

M\&M plain candies come in various colors. According to the M\&M/Mars Department of Consumer Affairs (link to the Mars company web site from the Brase/Brase statistics site at http://www.cengage.com/statistics /brase), the distribution of colors for plain M\&M candies is \begin{tabular}{l|ccccccc} \hline Color & Purple & Yellow & Red & Orange & Green & Blue & Brown \\ \hline Percentage & \(20 \%\) & \(20 \%\) & \(20 \%\) & \(10 \%\) & \(10 \%\) & \(10 \%\) & \(10 \%\) \\ \hline \end{tabular} Suppose you have a large bag of plain M\&M candies and you choose one candy at random. Find (a) \(P\) (green candy or blue candy). Are these outcomes mutually exclusive? Why? (b) \(P(\) yellow candy or red candy). Are these outcomes mutually exclusive? Why? (c) \(P(\) not purple candy \()\)
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