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Chapter 2: Problem 8

This is a continuation of Exercise 7. In many situations, the number of possibilities is not affected by order. For example, if a group of 4 people is selected from a group of 20 to go on a trip, then the order of selection does not matter. In general, the number \(C\) of ways to select a group of \(k\) things from a group of \(n\) things is given by $$ C=\frac{n !}{k !(n-k) !} $$ if \(k\) is not greater than \(n\). a. How many different groups of 4 people could be selected from a group of 20 to go on a trip? b. How many groups of 16 could be selected from a group of 20 ? c. Your answers in parts a and b should have been the same. Explain why this is true. d. What group size chosen from among 20 people will result in the largest number of possibilities? How many possibilities are there for this group size?

Short Answer

Expert verified
a) 4845 groups. b) 4845 groups. c) Choosing 4 people is the same as excluding 16. d) The largest number of combinations is for group size 10, with 184756 possibilities.

Step by step solution

01

Determine Combination Formula

We are given the formula for combinations as \( C = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. Combinations are used when the order of selection does not matter.
02

Solve for Part (a)

To find how many different groups of 4 people can be selected from a group of 20, set \( n = 20 \) and \( k = 4 \) in the formula. Thus, calculate \( C = \frac{20!}{4!(20-4)!} \). Simplify this to \( C = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} \), which equals \( 4845 \).
03

Solve for Part (b)

For selecting groups of 16 from 20, set \( n = 20 \) and \( k = 16 \) in the formula. This gives us \( C = \frac{20!}{16!(20-16)!} \). Simplify this to \( C = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} \), which also equals \( 4845 \).
04

Explain Part (c)

The answers to parts (a) and (b) are the same because selecting 4 people to go on a trip is equivalent to selecting the 16 people who will not go, from a total of 20 people. Mathematically, \( C(n, k) = C(n, n-k) \).
05

Determine Maximum Group Size for Part (d)

To find the group size that maximizes the number of combinations, observe that the combination formula is symmetric around \( n/2 \). Thus, the largest number of combinations occurs when \( k \) is about half of \( n \). For \( n = 20 \), the largest possibility occurs at \( k = 10 \). Calculate \( C = \frac{20!}{10!(20-10)!} \), which equals \( 184756 \).

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Key Concepts

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

Permutations
Permutations are all about arrangements where the order does indeed matter. Imagine a scenario where you have a deck of cards and you want to know how many ways you can arrange some or all of the cards in a particular order. That's where permutations sparkle. In essence, permutations relate to situations where the position or sequence of the selected items matters. Let's say you have the letters A, B, and C. If you want to know how many possible ways you can arrange two letters at a time from these, you'd use permutations. So, you would consider AB different from BA; both count as unique arrangements. The formula for permutations when choosing "r" items from "n" available is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \] Notice that it resembles the combinations formula, but without the division by "r!". This is because permutations worry about sequence, making them more numerous since each different sequence counts as a separate arrangement.
Factorials
Factorials are the building blocks of permutations and combinations. The concept of a factorial is quite simple: it's the product of all positive integers up to a certain number, denoted by the symbol "!". For example, the factorial of 4 is written as \(4!\) and calculated as \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] Factorials help in determining the total number of ways objects can be arranged. This arrangement calculation hinges on the idea that if you have "n" items, you have "n" choices for the first position, "n-1" for the second, and so on. Factorials are pivotal when figuring out combinations and permutations because they allow these calculations to be compactly expressed. Whenever you encounter a formula involving permutations or combinations, factorials are the mathematical magic behind the scenes working out how many ways you can mix and match items.
Binomial Coefficient
The binomial coefficient is a fascinating concept often described algebraically and combinatorially. It's a way to count combinations - that is, when order doesn't matter. This coefficient is generally notated as \( \binom{n}{k} \), a shorthand representation of \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, \(n\) represents the total number of items to choose from, and \(k\) denotes how many you want to choose. The binomial coefficient is an elegant way of expressing how many different groups can be formed when the arrangement within each group doesn't matter. For example, in the exercise above, choosing 4 people from 20 could be expressed using \( \binom{20}{4} \), showing the use of the binomial coefficient. Since it's symmetrical, choosing "k" items from "n" is the same as choosing "n-k" items, as evidenced by parts a and b of the exercise, which both resulted in 4845 combinations.

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Most popular questions from this chapter

The monthly profit \(P\) for a widget producer is a function of the number \(n\) of widgets sold. The formula is $$ P=-15+10 n-0.2 n^{2} . $$ Here \(P\) is measured in thousands of dollars, \(n\) is measured in thousands of widgets, and the formula is valid up to a level of 15 thousand widgets sold. a. Make a graph of \(P\) versus \(n\). b. Calculate \(P(1)\) and explain in practical terms what your answer means. c. Is the graph concave up or concave down? Explain in practical terms what this means. d. The break-even point is the sales level at which the profit is 0 . Find the break-even point for this widget producer.

In this exercise we develop a model for the growth rate \(G\), in thousands of dollars per year, in sales of a product as a function of the sales level \(s\), in thousands of dollars. \({ }^{30}\) The model assumes that there is a limit to the total amount of sales that can be attained. In this situation we use the term unattained sales for the difference between this limit and the current sales level. For example, if we expect sales to grow to 3 thousand dollars in the long run, then \(3-s\) gives the unattained sales. The model states that the growth rate \(G\) is proportional to the product of the sales level \(s\) and the unattained sales. Assume that the constant of proportionality is \(0.3\) and that the sales grow to 2 thousand dollars in the long run. a. Find a formula for unattained sales. b. Write an equation that shows the proportionality relation for \(G\). c. On the basis of the equation from part b, make a graph of \(G\) as a function of \(s\). d. At what sales level is the growth rate as large as possible? e. What is the largest possible growth rate?

The amount of vegetation eaten in a day by a grazing animal is a function of the amount \(V\) of food available (measured as biomass, in units such as pounds per acre). \({ }^{21}\) This relationship is called the functional response. If there is little vegetation available, the daily intake will be small, since the animal will have difficulty finding and eating the food. As the food biomass increases, so does the daily intake. Clearly, though, there is a limit to the amount the animal will eat, regardless of the amount of food available. This maximum amount eaten is the satiation level. a. For the western grey kangaroo of Australia, the functional response is $$ G=2.5-4.8 e^{-0.004 V} $$ where \(G=G(V)\) is the daily intake (measured in pounds) and \(V\) is the vegetation biomass (measured in pounds per acre). i. Draw a graph of \(G\) against \(V\). Include vegetation biomass levels up to 2000 pounds per acre. ii. Is the graph you found in part i concave up or concave down? Explain in practical terms what your answer means about how this kangaroo feeds. iii. There is a minimal vegetation biomass level below which the western grey kangaroo will eat nothing. (Another way of expressing this is to say that the animal cannot reduce the food biomass below this level.) Find this minimal level. iv. Find the satiation level for the western grey kangaroo. b. For the red kangaroo of Australia, the functional response is $$ R=1.9-1.9 e^{-0.033 V} $$ where \(R\) is the daily intake (measured in pounds) and \(V\) is the vegetation biomass (measured in pounds per acre). i. Add the graph of \(R\) against \(V\) to the graph of \(G\) you drew in part a. ii. A simple measure of the grazing efficiency of an animal involves the minimal vegetation biomass level described above: The lower the minimal level for an animal, the more efficient it is at grazing. Which is more efficient at grazing, the western grey kangaroo or the red kangaroo?

Friction loss in fire hoses: When water flows inside a hose, the contact of the water with the wall of the hose causes a drop in pressure from the pumper to the nozzle. This drop is known as friction loss. Although it has come under criticism for lack of accuracy, the most commonly used method for calculating friction loss for flows under 100 gallons per minute uses what is called the underwriter's formula: $$ F=\left(2\left(\frac{Q}{100}\right)^{2}+\frac{Q}{200}\right)\left(\frac{L}{100}\right)\left(\frac{2.5}{D}\right)^{5} $$ Here \(F\) is the friction loss in pounds per square inch, \(Q\) is the flow rate in gallons per minute, \(L\) is the length of the hose in feet, and \(D\) is the diameter of the hose in inches. a. In a 500 -foot hose of diameter \(1.5\) inches, the friction loss is 96 pounds per square inch. What is the flow rate? b. In a 500 -foot hose, the friction loss is 80 pounds per square inch when water flows at 65 gallons per minute. What is the diameter of the hose? Round your answer to the nearest \(\frac{1}{8}\) inch.

13\. Artificial gravity: To compensate for weightlessness in a space station, artificial gravity can be produced by rotating the station. \({ }^{8}\) The required number \(N\) of rotations per minute is a function of two variables: the distance \(r\) to the center of rotation, and \(a\), the desired acceleration (or magnitude of artificial gravity). See Figure \(2.58\) on the following page. The formula is $$ N=\frac{30}{\pi} \times \sqrt{\frac{a}{r}} $$ We measure \(r\) in meters and \(a\) in meters per second per second. a. First we assume that we want to simulate the gravity of Earth, so \(a=9.8\) meters per second per second. i. Find a formula for the required number \(N\) of rotations per minute as a function of the distance \(r\) to the center of rotation. ii. Make a graph of \(N\) versus \(r\). Include distances from 10 to 200 meters. iii. What happens to the required number of rotations per minute as the distance increases? Explain your answer in practical terms. iv. What number of rotations per minute is necessary to produce Earth gravity if the distance to the center is 150 meters? b. Now we assume that the distance to the center is 150 meters (so \(r=150\) ). i. Find a formula for the required number \(N\) of rotations per minute as a function of the desired acceleration \(a\). ii. Make a graph of \(N\) versus \(a\). Include values of \(a\) from \(2.45\) (one- quarter of Earth gravity) to \(9.8\) meters per second per second. iii. What happens to the required number of rotations per minute as the desired acceleration increases?
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