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In Problems 19-26, find the value of each combination. $$ { }_{8} C_{3} $$

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Step by step solution

01

Understand the Combination Formula

To find the value of the combination \(_{n}C_{r}\), the formula used is \(_{n}C_{r} = \frac{n!}{r!(n - r)!}\). In this case, n is 8 and r is 3.
02

Calculate the Factorials

Calculate the factorials involved. \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320\), \(3! = 3 \times 2 \times 1 = 6\), and \( (8 - 3)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
03

Substitute and Simplify

Substitute the calculated values into the combination formula: \(_{8}C_{3} = \frac{8!}{3!(8-3)!} = \frac{40320}{6 \times 120} \). Simplify the denominator first: \(6 \times 120 = 720\). Now, divide the numerator by the simplified denominator: \(\frac{40320}{720} = 56\).

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Key Concepts

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

Combination formula
When working with combinations in statistics, the combination formula is essential. Combinations help us determine the number of ways to choose a subset of items from a larger group, without regard to the order of selection. This formula is represented as \({ _{n}C_{r} }\), where \(n\) is the total number of items, and \(r\) is the number of items to choose.
To calculate this, we use the following formula: \[ _{n}C_{r} = \frac{n!}{r!(n - r)!} \] Applying this formula requires understanding and computing factorials for the values of \(n\) and \(r\).

For instance, in the problem \( _{8}C_{3} \), we can plug \(n = 8\) and \(r = 3\) into the formula to find our result. Remember: practice makes perfect! Try solving several exercises using this formula to build confidence and understanding.
Factorials
Factorials are fundamental when calculating combinations. A factorial, represented by the symbol \(!\), is the product of all positive integers up to a given number. For example, \(5!\) (5 factorial) is calculated as follows: \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] In our problem, we need to compute \(8!\), \(3!\), and \((8-3)!\).
  • \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320\)
  • \(3! = 3 \times 2 \times 1 = 6\)
  • \((8-3)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

Knowing how to compute factorials is key for simplifying the process of calculating combinations. It might look tough at first, but understanding each step makes it easier!
Probability concepts
Combinations play a crucial role in probability concepts. Probability measures the likelihood of an event occurring. When using combinations, we focus on how we can select items where the order does not matter. This distinction is important in probability calculations where the arrangement of items is irrelevant.

For example, if you want to find the probability of drawing 3 specific cards from a deck of 52 cards, you would use combinations, not permutations. Permutations consider the order of selection, while combinations do not. Using the combination formula, you can figure out how many possible ways there are to draw those 3 cards from the deck.
Combinations help simplify solving probability-related problems, making it an essential concept to grasp.
Statistical calculations
Statistical calculations often use combinations to solve real-world problems. In many fields like biology, finance, and computer science, understanding how to calculate combinations helps analyze data and make predictions.

In practice, you might use combinations to study experiments, surveys, or any scenario where choosing distinct groups matters. For example, choosing committee members from a larger pool or selecting a lottery ticket combination.
Consider the combination \( _{8}C_{3} \). By applying the combination formula, substituting \(n = 8\) and \(r = 3\), and computing the necessary factorials, the calculation process looks like this: \[ _{8}C_{3} = \frac{8!}{3!(8-3)!} = \frac{40320}{6 \times 120} = \frac{40320}{720} = 56 \] This result represents the ways to choose 3 items from 8, which simplifies complex analysis in various statistical applications.

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Most popular questions from this chapter

Suppose that two cards are randomly selected from a standard 52 -card deck. (a) What is the probability that the first card is a club and the second card is a club if the sampling is done without replacement? (b) What is the probability that the first card is a club and the second card is a club if the sampling is done with replacement?

Suppose that a digital music player has 13 tracks. After listening to all the songs, you decide that you like 5 of them. With the random feature on your player, each of the 13 songs is played once in random order. Find the probability that among the first two songs played (a) You like both of them. Would this be unusual? (b) You like neither of them. (c) You like exactly one of them. (d) Redo (a)-(c) if a song can be replayed before all 13 songs are played (if, for example, track 2 can play twice in a row).

Lingo In the gameshow Lingo, the team that correctly guesses a mystery word gets a chance to pull two Lingo balls from a bin. Balls in the bin are labeled with numbers corresponding to the numbers remaining on their Lingo board. There are also three prize balls and three red "stopper" balls in the bin. If a stopper ball is drawn first, the team loses their second draw. To form a Lingo, the team needs five numbers in a vertical, horizontal, or diagonal row. Consider the sample Lingo board below for a team that has just guessed a mystery word. $$ \begin{array}{|l|l|l|l|l|} \hline \mathbf{L} & \mathbf{I} & \mathbf{N} & \mathbf{G} & \mathbf{O} \\ \hline 10 & & & 48 & 66 \\ \hline & & 34 & & 74 \\ \hline & & 22 & 58 & 68 \\ \hline 4 & 16 & & 40 & 70 \\ \hline & 26 & 52 & & 64 \\ \hline \end{array} $$ (a) What is the probability that the first ball selected is on the Lingo board? (b) What is the probability that the team draws a stopper ball on its first draw? (c) What is the probability that the team makes a Lingo on their first draw? (d) What is the probability that the team makes a Lingo on their second draw?

A National Ambulatory Medical Care Survey administered by the Centers for Disease Control found that the probability a randomly selected patient visited the doctor for a blood pressure check is \(0.593 .\) The probability a randomly selected patient visited the doctor for urinalysis is 0.064. Can we compute the probability of randomly selecting a patient who visited the doctor for a blood pressure check or urinalysis by adding these probabilities? Why or why not?

The weather forecast says there is a \(10 \%\) chance of rain on Thursday. Jim wakes up on Thursday and sees overcast skies. Since it has rained for the past three days, he believes that the chance of rain is more likely \(60 \%\) or higher. What method of probability assignment did Jim use?

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