/*! 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 22 Find the value of each combinati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the value of each combination. $$_{12} C_{3}$$

Short Answer

Expert verified
220

Step by step solution

01

- Understand the Combination Formula

The combination formula to find the number of ways to choose k items from n items is \[ _{n}C_{k} = \frac{n!}{k!(n-k)!} \]where \( n \) is the total number of items, and\( k \) is the number of items to choose.
02

- Identify the Values of n and k

In this exercise, \( n = 12 \) and \( k = 3 \).
03

- Substitute the Values into the Formula

Substitute \( n \) and \( k \) into the combination formula as follows:\[ _{12}C_{3} = \frac{12!}{3!(12-3)!} \].
04

- Simplify the Factorials

Simplify the factorials:\[ 12! = 12 \times 11 \times 10 \times 9! \],\[ 3! = 3 \times 2 \times 1 \],and \[ (12-3)! = 9! \]. Notice that the \( 9! \) in the numerator and denominator cancel out.
05

- Perform the Calculation

Perform the remaining calculations: \[ _{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \]. Calculate the numerator: \[ 12 \times 11 \times 10 = 1320 \] and the denominator: \[ 3 \times 2 \times 1 = 6 \]. Now divide the results: \[ 1320 \div 6 = 220 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Combination Formula
To solve any problem involving combinations, we first need to understand the combination formula. This formula allows us to calculate the number of ways we can choose a subset of items from a larger set of items. The combination formula is given by: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)In this equation:
  • \(n\) is the total number of items.
  • \(k\) is the number of items we want to choose.
The exclamation mark (!) stands for factorial, which means multiplying a series of descending natural numbers.For instance, to find the number of ways to choose 3 items from a set of 12 items, we use \(n = 12\) and \(k = 3\). Placing these values into the formula, we get:\(\binom{12}{3} = \frac{12!}{3!(12-3)!}\).Next, we simplify the expression by substituting the values of the factorials.
Factorials
Factorials are essential for calculating combinations. A factorial, represented by an exclamation point (!), is the product of all positive integers up to a given number. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\) .In our exercise, we need to calculate factorials for 12 and 3, and simplify the expression: \(12! = 12 \times 11 \times 10 \times 9!\),\(3! = 3 \times 2 \times 1 = 6\), and \((12-3)! = 9!\).We notice that the \(9!\) terms cancel out each other, simplifying our calculation to: \(\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1}\).This cancellation reduces large numbers and makes the calculation easy to handle.
Probability Calculations
Finally, let's perform the actual calculations to find our combination. After cancelling out the factorial terms, our simplified combination formula is: \(\binom{12}{3} = \frac{12 \times 11 \times 10}{6}\).First, calculate the numerator: \(12 \times 11 \times 10 = 1320\).Next, calculate the denominator: \(3 \times 2 \times 1 = 6\).Now divide the numerator by the denominator: \(1320 \div 6 = 220\).So, the number of ways to choose 3 items from 12 items is 220.Understanding probability calculations helps in solving various problems in statistics and combinatorics, making these concepts very important.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Companies whose stocks are listed on the NASDAQ stock exchange have their company name represented by either four or five letters (repetition of letters is allowed). What is the maximum number of companies that can be listed on the NASDAQ?

According to the U.S Census Bureau, 19.1\% of U.S. households are in the Northeast. In addition, \(4.4 \%\) of U.S. households earn \(\$ 75,000\) per year or more and are located in the Northeast. Determine the probability that a randomly selected U.S. household earns more than \(\$ 75,000\) per year, given that the household is located in the Northeast.

In the U.S. Senate, there are 21 members on the Committee on Banking, Housing, and Urban Affairs. Nine of these 21 members are selected to be on the Subcommittee on Economic Policy. How many different committee structures are possible for this subcommittee?

How many different 10 -letter words (real or imaginary) can be formed from the letters in the word STATISTICS?

Suppose a computer chip company has just shipped 10,000 computer chips to a computer company. Unfortunately, 50 of the chips are defective. (a) Compute the probability that two randomly selected chips are defective using conditional probability. (b) There are 50 defective chips out of 10,000 shipped. The probability that the first chip randomly selected is defective is \(\frac{50}{10,000}=0.005=0.5 \% .\) Compute the probability that two randomly selected chips are defective under the assumption of independent events. Compare your results to part (a). Conclude that, when small samples are taken from large populations without replacement, the assumption of independence does not significantly affect the probability.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.