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Determine the value of each of the following using the appropriate formula. $$ \begin{array}{llllllllll} 3 ! & (9-3) ! & 9 ! & (14-12) ! & { }_{5} C_{3} & { }_{7} C_{4} & { }_{9} C_{3} & { }_{4} C_{0} & { }_{3} C_{3} & { }_{6} P_{2} & { }_{8} P_{4} \end{array} $$

Short Answer

Expert verified
The value for \(3!\) is 6, \((9-3)!\) is 720, \(9!\) is 362,880, \((14-12)!\) is 2, \( {5 \choose 3}\) is 10, \( {7 \choose 4}\) is 35, \( {9 \choose 3}\) is 84, \( {4 \choose 0}\) is 1, \( {3 \choose 3} \) is 1, \( {6P_2} \) is 30, \( {8P_4} \) is 1680.

Step by step solution

01

Factorial Calculation

To find the factorial of a number \( n \) is to multiply that number and all the fewer positive integers. Therefore, \( 3! = 3 \times 2 \times 1 = 6 \), \( (9-3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \), \( 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880 \) and \( (14-12)! = 2! = 2 \times 1 = 2 \).
02

Combination Calculation

The combination can be calculated by the formula \( {n \choose r} = \frac{n!}{r!(n-r)!} \), where \( n \) is the total items, and \( r \) is the items to be choosen. Thus, \( {5 \choose 3} = \frac{5!}{3!(5-3)!} = 10 \), \( {7 \choose 4} = \frac{7!}{4!(7-4)!} = 35 \), \( {9 \choose 3} = \frac{9!}{3!(9-3)!} = 84 \), \( {4 \choose 0} = \frac{4!}{0!(4-0)!} = 1 \), \( {3 \choose 3} = \frac{3!}{3!(3-3)!} = 1 \).
03

Permutation Calculation

The permutation can be calculated by the formula \( {nP_r} = \frac{n!}{(n-r)!} \), where \( n \) is the total items, and \( r \) is the items to be arranged. Thus, \( {6P_2} = \frac{6!}{(6-2)!} = 30 \) and \( {8P_4} = \frac{8!}{(8-4)!} = 1680 \).

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Key Concepts

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

Factorials
Factorials are a fascinating concept in mathematics. They are widely used in combinatorics, probability, and algebra. The factorial of a number, denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For instance, \( 3! = 3 \times 2 \times 1 = 6 \). It essentially represents the number of ways you can arrange \( n \) distinct objects.
Factorials grow very quickly as the numbers increase. For example, \( 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880 \). It is important to start the multiplication from the number down to 1.
An interesting special case is \( 0! \), which is defined as 1. This might seem counterintuitive at first, but it is important for mathematical consistency. Some aspects of probability and algebra would not work without this definition. In essence, factorials help in counting arrangements of items.
Combinations
Combinations allow us to choose a certain number of items from a larger set, without considering the order. This is crucial in scenarios where the order doesn’t matter, like forming a committee or selecting lottery numbers.
The formula for combinations is given by \( {n \choose r} = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items, and \( r \) is the number of items chosen. For example, choosing 3 items from a set of 5 can be calculated as \( {5 \choose 3} = \frac{5!}{3!(5-3)!} = 10 \).
Combinations have some interesting properties:
  • \( {n \choose r} = {n \choose n-r} \), meaning choosing \( r \) items from \( n \) is the same as choosing \( n-r \) items, since the leftover items form the remainder.
  • \( {n \choose 0} = 1 \) because there is exactly one way to choose nothing.
  • \( {n \choose n} = 1 \) because there is exactly one way to choose everything.
Overall, combinations are a key tool in problems where grouping elements are important, but sequence is not.
Permutations
Permutations help us figure out the number of possible arrangements of a set of items where the order does matter. Think about it as the possible ways to arrange books on a shelf or the different order arrangements in a queue.
The formula to determine permutations is \( P(n, r) = \frac{n!}{(n-r)!} \) where \( n \) is the total number of items and \( r \) is the number of items to arrange. For instance, the number of ways to arrange 4 out of 6 items can be calculated as \( 6P4 = \frac{6!}{(6-4)!} = 360 \).
Permutations emphasize:
  • Order is significant. Unlike combinations, changing the order of the chosen items results in a different permutation.
  • The number of permutations is always greater than or equal to the number of combinations for the same set of items.
This concept is widely applicable in situations where sequencing or prioritizing is important.

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Most popular questions from this chapter

Explain the meaning of the intersection of two events. Give one example.

Many states have a lottery game, usually called a Pick-4, in which you pick a four-digit number such as 7359 . During the lottery drawing, there are four bins, each containing balls numbered 0 through 9\. One ball is drawn from each bin to form the four-digit winning number. a. You purchase one ticket with one four-digit number. What is the probability that you will win this lottery game? b. There are many variations of this game. The primary variation allows you to win if the four digits in your number are selected in any order as long as they are the same four digits as obtained by the lottery agency. For example, if you pick four digits making the number 1265 , then you will win if \(1265,2615,5216,6521\), and so forth, are drawn. The variations of the lottery game depend on how many unique digits are in your number. Consider the following four different versions of this game. i. All four digits are unique (e.g., 1234 ) ii. Exactly one of the digits appears twice (e.g., 1223 or 9095 ) iii. Two digits each appear twice (e.g., 2121 or 5588 ) iv. One digit appears three times (e.g., 3335 or 2722 ) Find the probability that you will win this lottery in each of these four situations.

In how many ways can a sample (without replacement) of 9 items be selected from a population of 20 items?

Two thousand randomly selected adults were asked if they think they are financially better off than their parents. The following table gives the two- way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as. or worse off than their parents. $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Less Than } \\ \text { High School } \end{array} & \begin{array}{c} \text { High } \\ \text { School } \end{array} & \begin{array}{c} \text { More Than } \\ \text { High School } \end{array} \\ \hline \text { Better off } & 140 & 450 & 420 \\ \text { Same as } & 60 & 250 & 110 \\ \text { Worse off } & 200 & 300 & 70 \\ \hline \end{array} $$ Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. a. \(P\) (better off or high school) b. \(P\) (more than high school or worse off) c. \(P(\) better off or worse off \()\)

There is an area of free (but illegal) parking near an inner-city sports arena. The probability that a car parked in this area will be ticketed by police is \(.35\), that the car will be vandalized is \(.15\), and that it will be ticketed and vandalized is \(.10 .\) Find the probability that a car parked in this area will be ticketed or vandalized.

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