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Chapter 4: Problem 121

Find the value of each of the following using the appropriate formula. $$ \begin{array}{llllllllll} 6 ! & 11 ! & (7-2) ! & (15-5) ! & { }_{8} C_{2} & { }_{5} C_{0} & { }_{5} C_{5} & { }_{6} C_{4} & { }_{11} C_{7} & { }_{9} P_{6} & { }_{12} P_{8} \end{array} $$

Short Answer

Expert verified
Values: \(6! = 720\), \(11! = 39,916,800\), \((7-2)! = 120\), \((15-5)! = 3,628,800\), \(_{8}C_{2} = 28\), \(_{5}C_{0} = 1\), \(_{5}C_{5} = 1\), \(_{6}C_{4} = 15\), \(_{11}C_{7} = 330\), \(_{9}P_{6} = 60,480\), \(_{12}P_{8} = 199,584,000\)

Step by step solution

01

Calculate Factorials

To calculate \(6!\), multiply all positive integers from 1 to 6: \(6 * 5 * 4 * 3 * 2 * 1 = 720\). Similarly, to calculate \(11!\), multiply all positive integers from 1 to 11. For \((7-2)!\), first do the subtraction inside the brackets: \(7-2 = 5\). Then, calculate \(5!\). Lastly, calculate \((15-5)!\) in the same manner.
02

Calculate Combinations

To work out the combinations, use the combinatorial function formula which is: \[_{n}C_{r} = \frac{n!}{r!(n-r)!}\]When calculating \(_{8}C_{2}\), 'n' equals 8 and 'r' equals 2. Plug these values into the formula. Repeat this process for \(_{5}C_{0}\), \(_{5}C_{5}\), \(_{6}C_{4}\), and \(_{11}C_{7}\).
03

Calculate Permutations

To solve the permutations, the formula is:\[_{n}P_{r} = \frac{n!}{(n-r)!}\]Plug the values from \(_{9}P_{6}\) and \(_{12}P_{8}\) into this formula to get the answers.

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Key Concepts

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

Combinations
In mathematics, when we talk about combinations, we're discussing how to select items from a larger group. The key idea is that the order of selection doesn't matter. This means that choosing item A then B is the same as choosing item B then A. Combinations are determined using the formula: \[_{n}C_{r} = \frac{n!}{r!(n-r)!}\]Where:
  • \( n \) represents the total number of items.
  • \( r \) is the number of items to choose.
  • \(!\) denotes a factorial, which is the product of all positive integers up to that number.
For example, to calculate \(_{8}C_{2}\), you're finding the number of ways to choose 2 items from a set of 8, which is done by plugging \( n = 8 \) and \( r = 2 \) into the formula. It's important to follow the order of operations carefully to ensure accuracy in your results.
Permutations
Permutations are all about arranging items, where the order does matter. Unlike combinations, the sequence in which you arrange items is crucial.The formula for permutations is:\[_{n}P_{r} = \frac{n!}{(n-r)!}\]In this formula:
  • \( n \) is the total number of items.
  • \( r \) is the number of items to arrange.
  • \(!\) stands for factorial, which multiplies a series of descending natural numbers.
Consider \(_{9}P_{6}\) as an example. It calculates the number of ways to arrange 6 items out of 9. Here, the order of arrangement changes the final outcome, which is why it differs from combinations.
Combinatorial Formulas
Combinatorial formulas help us simplify complex calculations involving arrangements and selections. These formulas include those for both combinations and permutations.The two major formulas are:
  • The combinations formula: \[_{n}C_{r} = \frac{n!}{r!(n-r)!}\]
  • The permutations formula: \[_{n}P_{r} = \frac{n!}{(n-r)!}\]
Applying these formulas correctly allows you to solve problems related to counting and probability. They are essential tools in fields such as statistics and computer science. When using these formulas, it鈥檚 vital to understand the context: whether order matters (use permutations) or doesn鈥檛 matter (use combinations).Breaking down the expression step by step is crucial to avoid errors, especially with factorials and large numbers where calculations can become intricate.
Sequential Calculations
Sequential calculations refer to solving problems by breaking them down into a sequence of simpler calculations. This approach is particularly useful when dealing with factorials, combinations, and permutations.For instance, when solving \(6!\), you perform a series of multiplications: \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\). This is done step by step to manage and verify each part of the calculation.Similarly, when applying the combinations or permutations formula, it's helpful to follow a structured sequence:
  • Calculate the factorial of \( n \) or \( r \) as needed.
  • Simplify expressions inside parentheses first, if present.
  • Perform all multiplications and divisions in their intended order.
Sequential calculations enhance clarity and reduce mistakes, making it easier to tackle larger or more complex mathematical problems methodically.

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Most popular questions from this chapter

A production system has two production lines; each production line performs a two-part process, and each process is completed by a different machine. Thus, there are four machines, which we can identify as two first-level machines and two second-level machines. Each of the first-level machines works properly \(98 \%\) of the time, and each of the second-level machines works properly \(96 \%\) of the time. All four machines are independent in regard to working properly or breaking down. Two products enter this production system, one in each production line. a. Find the probability that both products successfully complete the two-part process (i.e., all four machines are working properly). b. Find the probability that neither product successfully completes the two- part process (i.e., at least one of the machines in each production line is not working properly).

A box contains 10 red marbles and 10 green marbles. a. Sampling at random from this box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time? b. Sampling at random from this box five times without replacement, you have drawn a red marble all five times. Without replacing any of the marbles, what is the probability of drawing a red marble the sixth time? c. You have tossed a fair coin five times and have obtained heads all five times. A friend argues that according to the law of averages, a tail is due to occur and, hence, the probability of obtaining a head on the sixth toss is less than \(.50 .\) Is he right? Is coin tossing mathematically equivalent to the procedure mentioned in part a or the procedure mentioned in part b above? Explain.

A screening test for a certain disease is prone to giving false positives or false negatives. If a patient being tested has the disease, the probability that the test indicates a (false) negative is \(.13 .\) If the patient does not have the disease, the probability that the test indicates a (false) positive is .10. Assume that \(3 \%\) of the patients being tested actually have the disease. Suppose that one patient is chosen at random and tested. Find the probability that a. this patient has the disease and tests positive b. this patient does not have the disease and tests positive c. this patient tests positive d. this patient has the disease given that he or she tests positive (Hint: A tree diagram may be helpful in part c.)

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

Given that \(P(B \mid A)=.70\) and \(P(A\) and \(B)=.35\), find \(P(A)\).
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