3.2: Counting Strategies
- Use tree diagrams to organize outcomes in a series of activities
- Develop and use the Multiplication Rule for counting the number of possible outcomes, including the factorial method
- Develop and use the permutation method for counting the number of possible outcomes in ordered arrangements
- Develop and use the combination method for counting the number of possible outcomes in unordered arrangements
Outcomes from a Series of Activities
We have discussed the classical method of computing probabilities for an event \(A\) by the formula \[ \nonumber P\left( A \right) = \frac{\text{number of ways event }A \text{ can occur}}{\text{number of outcomes in the sample space}} = \frac{x}{n}, \]provided outcomes in the sample space are equally likely. The formula requires determining the total number of outcomes in the sample space and in \(A.\) Up to this point, our examples and exercises have dealt with reasonably small sample spaces; however, the number of outcomes of interest may be quite large. This section develops counting techniques that help us in determine the number of possible outcomes. We begin by finding a logical way to organize the development of a sample space for multi-trial situations, such as tossing multiple dice. Ultimately, we will develop several counting methods.
Tree Diagrams and the Multiplication Rule for Counting
A tree diagram helps us list the various outcomes in a series of activities. To build a tree diagram, we list all outcomes of the first activity. Next to each of those listed outcomes, we branch to list all outcomes of the second activity. We branch off each of the second listed outcomes for the third activity, and so on. Eventually, all activities in the situation will be completed. Let us use a familiar example to build our first tree diagram. Suppose we wish to develop the sample space for rolling two dice. First, we list all the outcomes that may occur with the first die roll (the first activity of the described situation).
Figure \(\PageIndex{1}\): Tree diagram of first die roll
Next, from each of those six listed outcomes, we list all the outcomes that may occur with the second die roll (the second activity of the described situation).
Figure \(\PageIndex{2}\): Tree diagram of rolling two dice
Notice that, at the right of the diagram, we listed all thirty-six branches of the tree. This gives us an organized listing of the entire sample space. This branching idea hints at a counting strategy to determine the number of objects in the sample space. There are six branches for the first die roll and six branches from each for the second die roll; this leads to a total of \(6 \cdot 6\) \(=36\) outcomes in the sample space. This strategy is called the Multiplication Rule for Counting : if there are \(m\) possible outcomes for a first activity \(A_1\) and there are \(n\) possible outcomes for a second activity \(A_2,\) there are a total of \(m \cdot n\) possible outcomes for the experiment in which activity \(A_1\) is followed by activity \(A_2.\) This rule extends for any finite number of steps in a situation.
Create tree diagrams to determine the entire sample space of the following situations. Be sure to list the final sample space and then note the number of outcomes in the sample space.
- A food truck allows customers to create a meal by selecting one item from each category: burger, side, and drink. There are three burgers to choose from \( \left( B_1, B_2, B_3 \right),\) two sides \( \left( S_1, S_2 \right),\) and three drinks \( \left( D_1, D_2, D_3 \right).\) Create the sample space of different meals that can be ordered from this food truck.
- Answer
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We create the tree diagram as shown below. Notice that our first level of branches are the three burger choices, the second level of branches are the two side choices, and the third level of branches are the three drink choices:
Figure \(\PageIndex{3}\): Tree diagram of food truck meals
Notice that our diagram gives us our sample space. \begin{Bmatrix} B_1\sim S_1\sim D_1,~B_1\sim S_1 \sim D_2,~B_1\sim S_1\sim D_3, \\ B_1\sim S_2\sim D_1,B_1\sim S_2\sim D_2,~B_1\sim S_2\sim D_3, \\B_2\sim S_1\sim D_1,~B_2\sim S_1 \sim D_2,~B_2\sim S_1\sim D_3, \\ B_2\sim S_2\sim D_1,B_2\sim S_2\sim D_2,~B_2\sim S_2\sim D_3, \\ B_3\sim S_1\sim D_1,~B_3\sim S_1 \sim D_2,~B_3\sim S_1\sim D_3, \\ B_3\sim S_2\sim D_1,B_3\sim S_2\sim D_2,~B_3\sim S_2\sim D_3 \nonumber\end{Bmatrix}The sample space consists of \(18\) total outcomes. The Multiplication Rule of Counting predicts the total number of outcomes to be \(3 \cdot 2 \cdot 3\) \(= 18.\)
- There are six tiles well-mixed in a bag. The tiles are identical except in color; three are red, two are white, and one is blue. We are to randomly draw two tiles from the bag, one at a time, without replacement. Produce the sample space of two tiles that can be drawn from the bag.
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Again, we can create the diagram, but we must think clearly about each step of tile selection since the choices available for the second tile selection depend on what happened in the first tile selection. Since the tiles are identical, we cannot distinguish between those tiles of the same color. We might draw any of the three colors in the first tile selection, but in the second tile selection, we cannot select blue if we have selected blue on the first draw. We can organize our listing to produce the following tree diagram.
 Notice how our Multiplication Rule of Counting must be carefully used because we cannot select the blue tile again. There were two colors (specifically red and white) that, if selected first, still allowed three colors to be chosen in the second selection. There was one color (specifically blue) that, if selected first, only allowed two colors to be chosen in the second selection. We can adjust the rule's application so that \(2 \dot 3\) \(+ 1 \cdot 2\) \(= 8 \) predicts the total number of outcomes in the sample space.
- A salesman must visit four important clients: Mr. White, Miss Scarlet, Professor Plum, and Mrs. Peacock. Determine the number of choices the salesman has to visit these clients.
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We create the tree diagram as shown below, which gets a bit large. (One thing we might consider is to give codes to the four clients, such as numbering them \(1\) to \(4\) and using the numbers instead of their longer names.) The tree is reasonably easy to produce if we maintain our organization while creating each branch level, one at a time.
Figure \(\PageIndex{5}\): Tree diagram of client visits
In this situation, we will not list the sample space of \( 24\) distinct outcomes since it can be seen in the last column of the tree diagram. Notice how our Multiplication Rule of Counting predicts the number of outcomes at \( 4 \cdot 3 \cdot 2 \cdot 1 = 24 \). We have \(4\) options for the first visit, \(3\) options for the second, \(2\) options for the third, and then only \(1\) options for the final visit.
In our examples of tree diagrams, we can begin counting the number of outcomes by using our multiplication rule, but we must consider what happens at each branching action of the tree. Once we become comfortable with the concept of the multiplication rule, we can more quickly compute the number of possible outcomes.
First Basic Counting Techniques
We have shown that the Multiplication Rule for Counting is a valuable tool for determining the number of possible outcomes in many situations. We now use this tool to develop general counting techniques without producing tree diagrams. As we develop various techniques, remember that all the methods are developed from the Multiplication Rule for Counting.
Suppose June buys a new cell phone and must choose a six-digit code to keep it secure when she is not using it. How safe is her phone from someone who randomly chooses six digits to unlock it? Knowing that June will select one, we need to determine the number of possible six-digit codes. The standard convention in choosing a code allows for digits to repeat. Later, we will see what happens if the code cannot use digits more than once. We can think through the branching of a tree diagram using a simple counting diagram to determine the number of outcomes possible, \begin{matrix}\nonumber \text{# of Possibilities}&\text{ ___ }& \text{ ___ } & \text{ ___ } & \text{ ___ } &\text{ ___ } & \text{ ___ }&\, \\ \text{Choices for} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} & 6^{th} & \text{Code digit position} \end{matrix} where the row of blanks represent the numbers to be chosen in the given position in the code. We only need to place the number of possibilities in each blank for each code position. Recall that there are ten digits, \(0\) to \(9,\) and with use of the multiplication rule, we have \begin{matrix} \nonumber \text{# of Possibilities}&\underline{\, 10\, }& \underline{\, 10\,} & \underline{\, 10\,} & \underline{\, 10\,} &\underline{\, 10\,} & \underline{\, 10\, }&\Rightarrow 10 \cdot 10\cdot 10\cdot 10\cdot 10\cdot 10=10^6=1,000,000 \\ \text{Position} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} & 6^{th} & \, \end{matrix}There are \(1,000,000\) unique codes that June could choose from. The probability of someone randomly guessing June's code is \(P(\small{\text{GUESSING }}\)\(\small{\text{CODE}})\normalsize\)\( = \frac{1}{1,000,000}\)\( = 0.000001.\) This event is highly unlikely; June should feel relatively safe about someone randomly guessing her code. We would not want to construct the sample space in this situation due to the number of outcomes possible, yet we can still reflect on the sample space and know how many codes are in the sample space.
In Text Exercise \(\PageIndex{1}.3\) regarding a salesman visiting four distinct clients, we can use a similar approach: \begin{matrix}\nonumber \text{# of Possibilities}&\text{ ___ }& \text{ ___ } & \text{ ___ } & \text{ ___ } &\, \\ \text{Choices for} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & \text{Client visit position} \end{matrix}where the row of blanks represent the order to visit the four clients. We know that we have four clients to choose from in the first position; once that choice is made, there are three clients left to choose from for the second position; similarly, there are just two clients left to choose for the third position; and finally, only one client will be left for the fourth position: \begin{matrix} \nonumber \text{# of Possibilities}&\underline{\, 4\, }& \underline{\, 3\,} & \underline{\, 2\,} & \underline{\, 1\,} &\Rightarrow 4 \cdot 3 \cdot 2 \cdot 1 =24 \\ \text{Position} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & \, \end{matrix}There are \(24\) unique orderings the salesman could choose for visiting the four clients. This last example is a multiplication pattern that frequently occurs in counting outcomes. The shortened notation for this computation is called the factorial notation \(4!.\) The \(!\) symbol is read "factorial", \(4!\) is read "four factorial", and \(4!\) means, computationally, \(4 \cdot 3\cdot 2\cdot 1.\) In general \(n!\) represents the product \(n(n-1)(n-2)\cdot\ldots\cdot 2\cdot 1.\) We assign \(0! = 1\) by special definition. Notice how quickly factorials increase in value; for example \(8!\) \(= 40,320\) and \(20!\) \(= 2.4329 \times 10^{18}.\) Try the text exercises below for more practice with counting the number of possible outcomes.
Determine the total number of outcomes possible for each of the given descriptions.
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We have five standard fair coins of different values: a penny, a nickel, a dime, a quarter, and a half-dollar. When tossed, each lands on heads (H) or tails (T). One way they might land, in order of increasing value, is HHTHT.
- How many ways can the coins land when tossed and placed in order of value, say from smallest value to largest?
- How many ways can we order the five different coins?
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Both parts of this question ask for a count, and we use the multiplication rule for each, though the results are different.
- We establish our simple counting diagram, remembering that the coins are placed in order of increasing value: \begin{matrix}\nonumber \text{# of Possibilities}&\text{ ___ }& \text{ ___ } & \text{ ___ } & \text{ ___ } & \text{ ___ } &\text{each coin's landing side} \\ \text{Position of Coins} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} &\text{by denomination} \end{matrix} then we notice that each trial has two possible outcomes, either H or T. Our counting diagram becomes \begin{matrix} \nonumber \text{# of Possibilities}&\underline{\,2\, }& \underline{\, 2\,} & \underline{\, 2\,} & \underline{\, 2\,} & \underline{\, 2\,} &\Rightarrow 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5 =32 \\ \text{Position} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} & \, \end{matrix}This tells us there are \(32\) possible head and tail arrangements.
- Next, we examine the number of ways the five distinct coins can be placed in an order, such as from smallest value to largest, largest to smallest, or even some other arrangement. It is important to note that each of the coins is distinguishable from the other, so there is no confusion on which coin is in which location. We establish our simple counting diagram as an aide: \begin{matrix} \nonumber \text{# of Possibilities} & \text{ ___ } & \text{ ___ } & \text{ ___ } & \text{ ___ } & \text{ ___ } &\, \\ \text{Position in Arrangement} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} &\text{by coin denomination} \end{matrix} and see there are five choices for a coin to be placed in the \(1^{st} \) position. Once that coin is chosen, we have only four choices for the coin in the second position, and so on. Our counting diagram becomes \begin{matrix} \nonumber \text{# of Possibilities}&\underline{\,5\, }& \underline{\, 4\,} & \underline{\, 3\,} & \underline{\, 2\,} & \underline{\, 1\,} &\Rightarrow 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1= 5! = 120 \\ \text{Position} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} & \, \end{matrix}This tells us there are \(120\) possible arrangements of the coins by denomination. In the upcoming optional section, we will revisit this situation, in which some of the coins are not distinguishable from each other, requiring adjustments in our counting method.
- There are two distinctly colored standard decks of playing cards (red and blue). How many different outcomes are possible if we draw four cards in sequence, with two cards drawn first from the red deck and then two more from the blue deck?
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We again establish our simple counting diagram: \begin{matrix}\nonumber \text{# of Possibilities}&\text{ ___ }& \text{ ___ } & \text{ ___ } & \text{ ___ } &\, \\ \text{Choices for} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} &\text{each card's position} \\ \, & \text{red } & \text{cards} & \text{blue} & \text{cards}& \, \end{matrix}then we determine the possible outcomes for each card position. Our counting diagram becomes \begin{matrix} \nonumber \text{# of Possibilities}&\underline{\,52\, }& \underline{\, 51\,} & \underline{\, 52\,} & \underline{\, 51\,} &\Rightarrow 52 \cdot 51 \cdot 52 \cdot 51 = 7,033,104 \\ \text{Position} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & \, \\ \, & \text{red } & \text{cards} & \text{blue} & \text{cards}& \, \end{matrix}That is over \(7\) million different possible outcomes! We are definitely glad we did not have to produce the entire sample space for this simple little situation.
- Leisha is a research biologist for a sod (grass) company. She must research the effects of four fertilizer types on three temperature zones with three watering amounts. How many different sod plots must she create to test all possible configurations of these three factors?
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We jump to the final computing action to tell Leisha how many plots she will need to perform her research: \begin{matrix}\nonumber \text{# of Possibilities}&\underline{\,4\, }& \underline{\, 3\,} & \underline{\, 3\,} &\Rightarrow 4 \cdot 3 \cdot 3 = 36 \\ \text{Factor Choice} & \text{fertilizer} & \text{temp. zones} & \text{water amts.} &\, \end{matrix}Leisha will need to have \(36\) different plots for her desired sod research.
- Carl is a psychologist studying telepathy between patients who claim such abilities. He will test their claims using a deck of six cards, each with a different picture. The cards will be shuffled randomly before each pair of patients uses the cards in his experiment. How many different card arrangements (shuffled decks) are possible?
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Again, we jump to the final computing action to answer Carl's dilemma:\begin{matrix}\nonumber \text{# of Possibilities}&\underline{\,6\, }& \underline{\, 5\,} & \underline{\, 4\,} & \underline{\, 3\,} & \underline{\, 2\,} & \underline{\, 1\,} &\Rightarrow 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 6! = 720 \\ \text{Card deck position} & 1^{st} & 2^{nd} & 3^{rd} & 4^{th} & 5^{th} & 6^{th} &\, \end{matrix}So Carl will have \(6!\) \(=720\) different card arrangements possible in the deck of six cards. We mention that this is a factorial count situation. Once we recognize a factorial count, we will reach the final answer of \(6!\) with no diagram work. We note for Carl that if the probability of one patient in the study randomly guessing all six cards correctly as examined only by their telepathy partner is \( \frac{1}{720}\) \(\approx 0.1389\%\)--not impossible but highly improbable.
Counting with Permutations
Suppose, in the last problem from Text Exercise \(\PageIndex{2},\) Carl selected only three of the six cards for his patients to test their telepathic abilities. We can still use the same method described above. \begin{matrix}\nonumber \text{# of Possibilities}&\underline{\,6\, }& \underline{\, 5\,} & \underline{\, 4\,} &\Rightarrow 6 \cdot 5 \cdot 4 = 120 \\ \text{Card deck position} & 1^{st} & 2^{nd} & 3^{rd} &\, \end{matrix}Carl will have \(120\) different smaller deck arrangements if only selecting any three of the six original cards for the smaller deck.
More formally, we call situations that are counted this way permutations. A permutation count is the number of ways to arrange, in order, \(n\) distinct objects, taking \(r\) at a time. We denote this symbolically as \( \sideset{_{n}}{_{r}}P\) where \(n\) and \(r\) are whole numbers with \(n \ge r.\) The two general forms for computing this count are given by:\[ \begin{align*}\hspace{1in}\sideset{_{n}}{_{r}}P &=n(n-1)(n-2) \cdots (n-r+1) \hspace{1in}&&(1)\\&=\dfrac{n!}{(n-r)!}&&(2) \end{align*}\]When building the counting action, we tend to use \((1).\) When working with permutations theoretically, we tend to represent permutation counts using \((2).\) We should become familiar with both forms of computing permutations. Using Carl's smaller deck we can see the computation as \[ \nonumber \sideset{_6}{_3}P =n(n-1)(n-2) \cdots (n-r+1)= 6 \cdots (6 - 3 +1) = 6 \cdot 5 \cdot 4=120 \\ \text{and also} \\ \sideset{_{6}}{_{3}}P =\frac{n!}{(n-r)!}= \frac{6!}{(6-3)!}=\frac{6 \cdot 5 \cdot 4 \cdot \cancel{3} \cdot \cancel{2} \cdot \cancel{1}}{\cancel{3} \cdot \cancel{2} \cdot \cancel{1}}=120\]Our fraction form of computation with factorials is canceling out the unused \((n-r)!\) objects in the count from the \(n!.\) We also want to quickly recognize situations requiring permutation counts (especially when large numbers are involved) so we can use technology to perform the calculations.
Determine the number of outcomes possible in the described situations:
- Matching questions are sometimes used in statistics exams. If ten distinct statistical symbols are to be matched uniquely with ten distinct statistical terms, how many symbols-to-terms matches (correct or incorrect) are possible?
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We could use a counting diagram, but we produce the counts by recognizing the strategies discussed without using a diagram in these answers. Thinking about our statistical term choices for each of the ten symbols, we see there are \(10 \cdot 9 \cdot 8 \cdots 2 \cdot 1\) \(=10!\) \(= 3,628,800\) different symbol-to-term matches students can make. A student strictly randomly guessing all ten correctly has a probability of only \(\frac{1}{3,628,800}\) \(\approx 0.0000002756.\) We recognize that this count is not only a simple factorial count but that any factorial count can be considered a permutation count. Specifically \(10!\) \(=\sideset{_{10}}{_{10}}P\) \(=\frac{10!}{(10-10)!}.\)
- In the same matching question used above, if there are only seven distinct statistical symbols to be matched with the given ten distinct statistical terms, how many symbols-to-terms matches might be made by students?
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In this case, we only match seven symbols (places) with ten terms (objects). There are \( 10 \cdot 9 \cdots 4\) \(= \sideset{_{10}}{_{7}}P\) \(= 604,800 \) different symbol-to-term matches that students can make. After a little thought, we should recognize that this is a permutation counting situation since the order is important.
- A saleswoman must visit only four of her nine clients tomorrow. How many different ordered client visit lists can the saleswoman make?
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At this point, we recognize quickly that this is also a permutation count since order matters. The saleswoman has \( \sideset{_{9}}{_{4}}P\) \(= 3,024 \) ordered client visit lists that she might make. We doubt she will make all these, but knowing how many are possible speaks to the variety of ordered lists she has available. We can also use technology to compute such permutation values after we note the values of \(n\) and \(r.\) For example, in an Excel spreadsheet cell, we can enter \(=\text{PERMUT}(9, 4)\) and the value \(3024\) will be shown.
- During a two-hour time frame, a nurse must take vital signs measurements from ten of his forty assigned patients. How many different lists can the nurse make for an ordered round involving ten of the forty patients?
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We quickly recognize that this is also a permutation count. The nurse has \( \sideset{_{40}}{_{10}}P \) ordered patient lists that he might make for his two-hour round. We will use technology for the computation: in an Excel spreadsheet cell, we can enter \(=\text{PERMUT}(40,10)\) producing the approximate value \(3.07599\)E+\(15;\) this is a standard technology representation of the number \( 3.07599 \times 10^{15}.\) He can make well over a quadrillion ordered patient lists in this situation. We might be amazed at the possibilities and appreciate how many randomly chosen ordered lists can be made.
- We return to our "Birthday" problem of Text Exercise \(3.1.5,\) which asks, "What is the probability that in a random group of thirty-five people, two or more of the group will share a birth day in the year (such as September \(1^{st}\) or May \(28^{th}\))?" To answer this question, we assume \(365\) days in a year (ignore leap year cases), that each day of the year is equally likely for births, and the birth day of the various individuals in the group is not dependent on another individual in the group.
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We first note that in a group of thirty-five people placed in some order, the first person will have one of \(365\) different birth days, the second will have one of \(365\) birth days, ... ,the thirty-fifth will have one of \(365\) birth days. There are \( 365^{35}\) \(\approx 4.7891 \times 10^{89} \) possible lists of thirty-five birth days that various groups of thirty-five people can form.
Now, we must find how many of those lists have two or more with the same birth day in the year. To do so, we must look at many separate cases: how many of our lists have exactly two in a birth day list the same or how many have precisely three the same or ... or how many have all thirty-five with the same birth day or how many have two different pairs the same or ...etc. But being reflective on the work in finding so many separate counts and being wise to consider other methods, we again notice that there is only one case in the complement description: having lists in which none of the thirty-five have the same birth date. The use of the complement method makes this problem approachable. Therefore, we must count only the number of lists of birth days that can be made in which none of the thirty-five are the same birth day.
Since this is an ordered list, we recognize this is a permutation count in which we are forming lists of \(35\) distinct birth days coming from \(365\) options...that is, we need \( \sideset{_{365}}{_{35}}P\) \(= \frac{365!}{(365 - 35)!}\) \(\approx 8.8893 \times 10^{88}.\) So, \(P(\small{\text{NONE }}\)\(\small{\text{HAVE }}\)\(\small{\text{THE }}\)\(\small{\text{SAME }}\)\(\small{\text{BIRThDAY}}\normalsize)\) \( \approx \frac{8.8893 \times 10^{88}}{4.7891 \times 10^{89}}\) \( \approx 0.1856.\)
Now, the \(P (\small{\text{AT }}\)\(\small{\text{LEAST }}\)\(\small{\text{TWO }}\)\(\small{\text{WITH }}\)\(\small{\text{SAME }}\)\(\small{\text{BIRTHDAY}})\)\(=1-P(\small{\text{NONE }}\)\(\small{\text{HAVE }}\)\(\small{\text{THE }}\)\(\small{\text{SAME }}\)\(\small{\text{BIRTHDAY}})\) \( \approx 1 - 0.1856\) \(=0.8142\) \(=81.42\%.\) Notice that in a randomly selected group of thirty-five people, there is a reasonably high probability that at least two will share the same birth day.
It is important to note that order is essential in permutations. Choosing object \(a\) and then object \(b,\) in that order, is counted as a different outcome from choosing object \(b\) first and then object \(a.\) When the order of choice is irrelevant (such as the saleswoman choosing to visit four of her nine clients but not concerned with the order of those of her visits), an adjusted approach must be taken: a counting approach termed as a combination.
Counting with Combinations
In each of our previous counting approaches, the order of selection was important to the counting. However, there are times when order is not important to the outcomes of an experiment, especially when we consider samples taken from a population. We begin with a simple, small example that we will later generalize.
Suppose we have four distinct individuals, Adam \( ( \text{A} ),\) Betsy \( ( \text{B} ),\) Cathy \( ( \text{C} ),\) and Damon \( ( \text{D} )\) and we wish to form all possible sample groups of size three from this population of four. Using our tree approach, we can create the \( \sideset{_{4}}{_{3}}P\) \( = 24 \) ways to arrange these four into ordered groups of three--yes, a permutation count. This list is given by: \begin{matrix} \nonumber \color{red} \text{ABC} & \color{blue} \text{DBC} & \color{gray}\text{ACD} & \text{ABD} \\ \color{red}\text{ACB} & \color{blue}\text{DCB} & \color{gray}\text{ADC} & \text{ADB} \\ \color{red}\text{BCA} & \color{blue}\text{BCD} & \color{gray}\text{CDA} & \text{BDA} \\ \color{red}\text{BAC} & \color{blue}\text{BDC} & \color{gray}\text{CAD} & \text{BAD} \\ \color{red}\text{CAB} & \color{blue}\text{CDB} & \color{gray}\text{DAC} & \text{DAB} \\ \color{red}\text{CBA} & \color{blue}\text{CBD} & \color{gray} \text{DCA} & \text{DBA} \end{matrix}The order displayed here is intentionally a bit different from our standard tree approach. Thinking back on our need to count how many sample groups of size three are available, we should note that the various colored collections of the ordered groups produce the same (unordered) sample groups. The first column of six shown in red is different arrangements of the same sample group containing Adam, Betsy, and Cathy; the second column of six in blue contains only one sample group of Betsy, Cathy, and Damon; the third column has the one sample group of Adam, Cathy, and Damon; and the fourth column of Adam, Betsy, and Damon. Given any collection of three individuals, we know there are \(3 \cdot 2 \cdot 1\) \(= 6\) ways to arrange those three. There are only \( \frac{ \sideset{_{4}}{_{3}}P }{3!}\) \(= \frac{24}{6}\) \(= 4\) unique sample groups (unordered groups) of size three from this collection of four individuals.
Is there a predictable pattern to develop a counting formula for the number of distinct unordered groups of size \(r\) taken from a collection of \(n\) distinct objects? First, we performed a permutation count of \( \sideset{_{n}}{_{r}}P \) to get the total number of ordered groups of size \(r\) taken from the \(n\) objects. Then, we had to divide by the number of ways any specific group of size \(r\) could be arranged--a factorial value of \(r!\)--to account for the same collection arranged differently. We computed \( \frac{ \sideset{_{n}}{_{r}}P }{r!} \) to determine the number of distinct combinations of \(n\) distinct objects taken \(r\) at a time. We call this type of counting a combination and denote in symbols as \( \sideset{_{n}}{_{r}}C \) or as \( \left( \begin{matrix} n \\ r \end{matrix} \right) \); in this text, we will use the \( \sideset{_{n}}{_{r}}C \) form for consistency. We typically read this as, "\(n\) choose \(r\)" since it is counting how many ways we can choose \(r\) objects from a group of \(n\) objects. We can express the formula for combination count in two ways.\[ \nonumber \left( \begin{matrix} n \\ r \end{matrix} \right) = \sideset{_{n}}{_{r}}C = \dfrac{ \sideset{_{n}}{_{r}}P }{r!}=\dfrac{n!}{(n-r)!r!}\]We can reason algebraically how the second formula is produced from the first by remembering that \( \sideset{_{n}}{_{r}}P\) \(= \frac{n!}{(n-r)!}.\) Many computing technologies also have built-in functions to relieve us from mechanical computation once we have recognized a counting situation as a combination. In the Excel spreadsheet, this function is \(=\text{COMBIN}(n,r).\)
We must remember that a combination counts the number of ways to form groups of size \(r\) from a collection of size \(n\) where order within the groups does not matter. Let us do one more example before a collection of text exercises. Suppose a committee of four students must be formed from a class of thirty-five distinct students. How many committees could be formed? Rearranging the order of four students does not create a different committee. Therefore, we must count using combinations. There are \( \sideset{_{35}}{_{4}}C\) \(= \frac{ \sideset{_{35}}{_{4}}P }{4!}\) \(=\frac{35 \cdot 34 \cdot 33 \cdot 32}{4 \cdot 3 \cdot 2 \cdot 1}\) \(= 52,360\) different possible committees of size four that could be formed from the thirty-five students. Using the Excel command \(=\text{COMBIN}(35,4)\) produces this same value of \(52,360.\)
Determine the number of outcomes possible in the described situations (warning: not all are combination counts).
- A construction manager must select eight samples from a group of twenty concrete trucks to test the quality of the concrete mixture. How many different groups of eight can be selected?
- Answer
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Notice that initially, there are twenty trucks to choose between for the first sample, nineteen for the second, and so on until all eight needed trucks are chosen. However, we also recognize that once a group of eight trucks is selected in a specific order, rearranging those eight trucks in a different order will not create a different sample group. This is a combination counting situation, so we compute \(20\) choose \(8:\)\[ \nonumber \sideset{_{20}}{_{8}}C =\frac{20!}{(20-8)! \cdot 8!} = \frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}{ 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 125,970.\]There are almost \(126,000\) sample groups of concrete trucks possible for the construction manager to choose. We could also have computed our value in a spreadsheet using \(=\text{COMBIN}(20,8)\) in a cell, producing this same end value of \(125,979.\)
- A dial-faced combination lock produced by a new manufacturer uses three distinct numbers in the lock code between \(0\) and \(39.\) How many different lock codes are available?
- Answer
-
In this case, we are choosing three numbers from a list of forty\( (0\) to \(39).\) We also recognize that once a group of three numbers is chosen, rearranging those three numbers will create a different lock code. This is a permutation counting situation, so we compute \(40\) permute \(3\):\[ \nonumber \sideset{_{40}}{_{3}}P =\frac{40!}{(40-3)!} = 40 \cdot 39 \cdot 38 = 59,280.\]There are \(59,280\) lock codes for the described lock. We should notice that the term "combination lock" does not use the term "combination" in the same fashion as our counting method; that is, the number of unlock codes for a combination lock is not determined by a combination counting strategy. Others' general use of the term "combination" may not align with our use of the term in counting strategies. We also note the computation in a spreadsheet using \(=\text{PERMUT}(40,3)\) in a cell produces this same computed value of \(59,280.\) For the remainder of our examples, we will use the spreadsheet for computation work after determining the type of counting method needed.
-
A qualified applicant pool for a large hospital's eight nurse trainee positions consists of ten women and six men.
- How many different groups of these sixteen applicants can be formed for the eight positions?
- How many different groups can be selected from the pool of applicants that consist entirely of women?
- If all applicants are considered equally qualified, and the group of eight positions is randomly selected, what is the probability that the chosen group will consist entirely of women? (This measure could be of interest in a "discrimination" claim if the chosen pool consisted only of women.)
- Answer
-
In this situation, we have sixteen applicants; eight are needed to form our trainee group.
- We recognize that we are choosing eight from our group of sixteen, and the order of arrangement of any group of eight will not make a different trainee group. This is a combination counting situation, so we compute \(16\) choose \(8.\)\[ \nonumber \sideset{_{16}}{_{8}}C = \text{COMBIN}(16,8)=12,870\]There are well over \(10,000\) trainee groups that can be formed from the pool of applicants.
- For the selected group to be all women, we recognize that we are choosing eight from our group of ten women. We again notice that the order of arrangement of any group of eight women will not make a different trainee group; the selection order does not matter. This is a combination counting situation; we compute \(10\) choose \(8.\)\[ \nonumber \sideset{_{10}}{_{8}}C = \text{COMBIN}(10,8)=45\]There are \(45 \) trainee groups that can be formed only from the pool of women applicants.
- Assuming random selection in which any group of eight is equally likely to be chosen, the probability that the selected group will consist entirely of women is \( P(\small{\text{ALL }}\)\(\small{\text{WOMEN}}\normalsize)\) \(=\frac{45}{12,870}\) \( \approx 0.3497\%,\) which we would consider an unusual event.
-
To win a specific
Powerball
lottery grand prize, a player must match all \(5\) white balls (order does not matter) and the red Powerball on his purchased ticket.
- How many different tickets could be purchased for entry into this type of Powerball lottery game?
- Knowing there is only one winning ticket per game, what is the probability that a randomly generated ticket will be the winning ticket for the grand prize?
- Answer
-
Notice that initially, we can see two steps in this activity of completing an entry to the Powerball game. We choose five numbers from a list of sixty-nine and then select one number from a list of twenty-six. We use our multiplication rule between the counts of the two steps.
- For the first step, we choose any five of the sixty-nine white-ball numbers, recognizing that once a specific group of five numbers is selected, rearranging those five chosen numbers will not make a different Powerball ticket. This is a combination counting situation, so we compute \( \sideset{_{69}}{_{5}}C\) \(= 11,238,513 .\) For the second step, we choose any of the twenty-six red-ball numbers, with a count of \(26\) possible choices. (We could do a combination count of \( \sideset{_{26}}{_{1}}C\) \(= \text{COMBIN}(26,1)\) as well, but this still produces that same value of \(26\) and involves unnecessary computation work. Thinking as we make our counts is more valuable than computation work.) There are a total of \( \sideset{_{69}}{_{5}}C \cdot 26\) \9= 292,201,338 \) possible tickets that could be made for any single Powerball lottery game.
- Assuming a random selection of numbers for the winning ticket, the probability that a randomly generated ticket will be the winning ticket is \( P(\small{\text{WINNING }}\)\(\small{\text{TICKET}}\normalsize)\) \(=\frac{1}{292,201,338}\) \( \approx 3.4223 \times 10^{-9}.\) As we are probably all aware, winning a Powerball Lottery game by purchasing a single ticket is a highly unusual event. With these large numbers, we can also reason that we cannot create and purchase all possible tickets to guarantee winning.
Section Summary
This section on counting has been extensive. As a final review, let us list our various counting strategies.
- Tree diagram (useful for creating and listing all possible outcomes for a situation in which multiple steps are used to produce the branches of the trees).
- Multiplication Rule for Counting
- Factorial Rule for Counting
- Permutation Rule for Counting
- Combination Rule for Counting
Knowing which counting method applies requires us to think clearly about the selection of the objects in steps and to determine if the order of selection matters to the count.


2.4B: Types of Biological Macromolecules
Biological macromolecules, the large molecules necessary for life, include carbohydrates, lipids, nucleic acids, and proteins.
- Identify the four major classes of biological macromolecules
Key Points
- Biological macromolecules are important cellular components and perform a wide array of functions necessary for the survival and growth of living organisms.
- The four major classes of biological macromolecules are carbohydrates, lipids, proteins, and nucleic acids.
Key Terms
- polymer : A relatively large molecule consisting of a chain or network of many identical or similar monomers chemically bonded to each other.
- monomer : A relatively small molecule that can form covalent bonds with other molecules of this type to form a polymer.
Nutrients are the molecules that living organisms require for survival and growth but that animals and plants cannot synthesize themselves. Animals obtain nutrients by consuming food, while plants pull nutrients from soil.
Sources of biological macromolecules : Foods such as bread, fruit, and cheese are rich sources of biological macromolecules.
Many critical nutrients are biological macromolecules. The term “macromolecule” was first coined in the 1920s by Nobel laureate Hermann Staudinger. Staudinger was the first to propose that many large biological molecules are built by covalently linking smaller biological molecules together.
Living organisms are made up of chemical building blocks : All organisms are composed of a variety of these biological macromolecules.
Monomers and Polymers
Biological macromolecules play a critical role in cell structure and function. Most (but not all) biological macromolecules are polymers, which are any molecules constructed by linking together many smaller molecules, called monomers. Typically all the monomers in a polymer tend to be the same, or at least very similar to each other, linked over and over again to build up the larger macromolecule. These simple monomers can be linked in many different combinations to produce complex biological polymers, just as a few types of Lego blocks can build anything from a house to a car.
Monomers and polymers : Many small monomer subunits combine to form this carbohydrate polymer.
Examples of these monomers and polymers can be found in the sugar you might put in your coffee or tea. Regular table sugar is the disaccharide sucrose (a polymer), which is composed of the monosaccharides fructose and glucose (which are monomers). If we were to string many carbohydrate monomers together we could make a polysaccharide like starch. The prefixes “mono-” (one),”di-” (two),and “poly-” (many) will tell you how many of the monomers have been joined together in a molecule.
The molecule sucrose (common table sugar) : The carbohydrate monosaccharides (fructose and glucose) are joined to make the disaccharide sucrose.
Biological macromolecules all contain carbon in ring or chain form, which means they are classified as organic molecules. They usually also contain hydrogen and oxygen, as well as nitrogen and additional minor elements.
Four Classes of Biological Macromolecules
There are four major classes of biological macromolecules:
- carbohydrates
- lipids
- proteins
- nucleic acids
Each of these types of macromolecules performs a wide array of important functions within the cell; a cell cannot perform its role within the body without many different types of these crucial molecules. In combination, these biological macromolecules make up the majority of a cell’s dry mass. (Water molecules make up the majority of a cell’s total mass.) All the molecules both inside and outside of cells are situated in a water-based (i.e., aqueous) environment, and all the reactions of biological systems are occurring in that same environment.
Interactive: Monomers and Polymers : Carbohydrates, proteins, and nucleic acids are built from small molecular units that are connected to each other by strong covalent bonds. The small molecular units are called monomers (mono means one, or single), and they are linked together into long chains called polymers (poly means many, or multiple). Each different type of macromolecule, except lipids, is built from a different set of monomers that resemble each other in composition and size. Lipids are not polymers, because they are not built from monomers (units with similar composition).


10.4: VSPER Theory - Predicting Molecular Geometries
enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible. Valence shell electron-pair repulsion theory ( VSEPR theory)
VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.
As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF 2 molecule. The Lewis structure of BeF 2 (Figure \(\PageIndex{2}\)) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure \(\PageIndex{2}\)).
Figure \(\PageIndex{2}\): The BeF 2 molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom.
Figure \(\PageIndex{3}\) illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.


13.6: Credit Ratings
Corporations are rated by credit-rating agencies that assess a corporation’s ability to service its debt (i.e., to pay interest and principal in full and on time) and, thus, to keep bankruptcy at bay. This is a risk, which may also be referred to as “default risk.”
We have already examined three solvency ratios which attempt to provide some insight into this risk. Clearly, there are many more tools and considerations that enter into the process of credit analysis and rating. Companies must pay for this service, but not all do so, as some bond issues are too small to justify the expense, while others may be sold directly to institutional investors and thus do not require a rating.
Today, there are two major rating agencies: Standard & Poor’s (S&P) and Moody’s. Fitch is still a third agency, but it does not enjoy the market presence of the others. In addition, there are numerous nationally recognized statistical rating organizations or “NRSRO’s” that perform similar functions (see: https://www.sec.gov/ocr/ocr-current-nrsros.html ). The agencies rate bonds on the scale illustrated below. As you will note, there are some differences in the rating scales, and, although not visible in the ratings themselves, the manners in which the agencies conduct their respective analyses and what they consider important also differ from one agency to the other. As a result, the agencies may not rate the same bond issuer the same.
Within each rating category, the agencies may append additional notation, such as “A-” in the case of S&P. This provides some further refinement to the ratings.
Note:
Ratings will clearly affect the bond’s yield – in inverse relation to the rating with lower ratings generally bringing higher yields. The better the bond’s initial rating, the lower the bond’s coupon rate of interest, and, of course, the lower the issuer’s cost of debt capital; this is also true for the bond’s subsequent secondary market yield, (a.k.a. yield-to-maturity). Do not confuse coupon and market yields; they are separate.
Agencies do not necessarily agree with one another; in some cases, the agencies may rate the same company somewhat differently. Interestingly, the bond market itself seems to understand what the true yield for a bond should be; indeed, yields often adjust long before a rating change (i.e., either an upgrade or a downgrade) announcement. Further, a bond’s rating is not necessarily “correct.”
In the case of municipal bonds, the agencies conduct similar analyses and provide similar ratings. The interpretation of municipal ratings, however, is somewhat different in practice, with municipal bonds less likely to default than their ratings might imply. In some instances, municipal bonds may be insured by private insurance companies, in most of which cases the bonds will therefore bear a “AAA” rating regardless of its intrinsic creditability. Most insurers are considered to be in the best financial health.
United States Treasury and Agency bonds are highly rated but are no longer perceived as being completely risk free. “Agencies” are considered slightly lower in quality than “Treasuries” and will therefore trade at very slightly higher yields. The markets “know” how to price their securities.
While the rating agencies do not provide the details of the manner in which they conduct their ratings, it is accepted that their formulae consist of a combination of the examination of the issuer’s financial health (based on its financial statements), its economic environment, and an examination of the bond’s covenants.
Covenants are agreed-upon terms to which the issuer assures it will adhere; in a way it is a restriction imposed by the lender in order to secure or strengthen his position. For instance, the issuer may assure the investors that its financial ratios will remain within certain parameters, or that it will not issue any further debt, which shall be senior to the issue at hand. In general, such negative covenants will also prohibit the firm from paying too much in dividends, selling or pledging assets to other lenders, maintaining collateral in good condition, or adding on more debt. Timely interest and principal payments wo uld certainly be included in this category. These assurances are spelled out in the bond i ndenture , a legal document in which all the terms of the bond issue are spelled out, including the bond’s maturity and coupon interest rate.
Although a bond issuer, or borrower, may pay its bond interest in full and on time, which is to say it is not in default , any violation of any of the bond indenture’s covenants may be interpreted as a technical default and result in a downgrading of the bond’s credit rating. The bond covenants are aimed at providing investors with reasonable assurances of the issuer’s fealty to its obligations under the bond, but not so much as to disenable it from running its business effectively. Failure to abide by the covenants is, thus, a negative circumstance .
The restrictiveness of the bond’s covenants will vary . S ecured, or asset-backed , securities will have less restrictive covenants than unsecured issues. Furthermore, lower rated, non-investment grade, bonds are likely to be more restrictive and more complex, and hence will require a more careful reading prior to purchase and investment. Evaluation of these risk ier securities’ covenants is relatively more important than in the case of investment grade bonds. Moody’s places some emphasis in the determination of its ratings for high-yield bonds on the issue’s covenants. High yield bonds tend to attract more sophisticated investors who are capable of studying the technical terms of the covenants; moreover, many such issues are placed privately among institutional investors.
Investment grade bonds will typically contain three standard covenants. First, there may be a covenant regarding “ liens, ” which assure the investor that no other party may obtain a “senior” (prior) claim to the asset (or collateral) subsequent to the bond’s issuance. Under the law, a lien is a claim by a person upon another person’s property for purposes of securing payment of a debt or the fulfillment of an obligation. If there is a prior lien on a property, no other person can subsequently make a claim on the same property. By manner of another example, when you buy a house, your attorney will conduct a “lien search,” to be sure that no one else owns the house except the seller with whom you are dealing.
Next, a covenant will protect against any reduction in seniority of the bond in case of a merger of the issuer with another corporation. This covenant will ensure that the bond is paid off first in any case.
Finally, another covenant may address the sale of an underlying asset – in the case of a secured bond. Obviously, should an asset underlying a collateralized bond be sold, the bond’s financial status will change markedly. Such assets ought not to be sold unless the bond is also redeemed.
There is no doubt that credit ratings are critical to pricing primary and secondary market issues. Ratings influence both the coupon rates of interest in the primary market, and market yields, or yields-to-maturity in the secondary markets.


Unit 2: Postcolonial Literature
Unit 2: Postcolonial Literature
The post-World War II era and postcolonialism largely overlap. World War II (1939-1945) created a particular
zeitgeist
(or a defining spirit of the time), which consequently shaped the literary works published immediately after the war. Polish catholic writer Tadeusz Borowski's short story "This Way for the Gas, Ladies and Gentlemen" (1946) is an example of post-World War II literature, which reflects the author's experience of having observed, and having been complicit with, the brutal workings of the Holocaust. The first-person narrator's sense of nausea and being stuck in the short story echoes the major sentiments of post-World War II literature.
The post-World War II era also coincides with the post-independence period. After World War II and during the second half of the twentieth century, many formerly colonized places began to gain independence. In this historical milieu, postcolonialism as an academic field developed in the late 1970s—markedly with the 1978 publication of Edward Said's seminal book
Orientalism
, in which Said examines how euro-centric cultural representations of the "Orient" reveal the West's biases, stereotypes, and/or fantasies about the "Orient." Postcolonialism reached further sophistication by notable postcolonial scholars such as Homi Bhabha and Gayatri Spivak. Unlike the term "post-independence," which focuses on the temporal and periodizing meaning of "post" as "after," the term "postcolonial" has now become more of a theoretical concept, concerning the study of the varied processes and effects of colonialism from the perspectives of the (formerly) colonized; in this sense, postcolonialism could apply to cultures before, during, and after the workings of colonialism. Nigerian novelist Chinua Achebe's
Things Fall Apart
(1958) is one of the best-known works of postcolonial literature. Beyond the scope of the works produced from the newly decolonized countries of Africa, Asia, and the Caribbean, postcolonialism—as it concerns colonial histories and the voices of the oppressed people—is also associated with Indigenous peoples' experiences around the globe. Some of Joy Harjo's poems in this volume concern colonial history and Native Americans' experiences.
The post-independence era and postcolonialism come with other related concepts and new global realties, such as cultural hybridity, immigration, diaspora, and globalization. Postcolonial literature is also often noted for demonstrating cultural hybridity in its style and theme and/or engaging with the dilemma of conflicting cultures. Achebe's works, for example, engage dilemmas that emerge at the crossroads of different cultures.
Written by Kyounghye Kwon


7.8: Review Questions
________ is like the origin story for a brand.
- problem-solution narrative
- purpose
- mission statement
- vision statement
Identify the correct terms used in this text for setting SMART goals.
- specific, measurable, attainable, relevant, time-bound
- specific, marketable, attainable, relevant, time-bound
- specific, measurable, achievable, relevant, timely
- specific, measurable, attractive, relevant, time-bound
What is defined as the thing your organization does that people will pay for or contribute to?
- earnings statement
- profit proposition
- value for revenue proposal
- value proposition
When an entrepreneur’s story becomes part of the mythology of a corporation and its purpose, this is known as a/an ________.
- mission statement
- vision statement
- purpose statement
- corporate narrative
When someone supports your products and your brand so much that they begin speaking on your behalf in person and on social media, they are known as a/an ________.
- brand advocate
- corporate narrative sponsor
- impetus
- entrepreneurial drive
In previous decades, a term for a related phenomenon was “Keeping up with the Joneses.” This refers to a sense that if we are not living our best life, having the most fun, enjoying the best foods, fashion, and so on, then we are not living the life we should. This term can refer to both a short-term concern and a general anxiety that others are enjoying life more than we are.
- brand advocacy awareness
- FOMO, or fear of missing out
- “Jonesing”
- happy ever anxiety
Pitches have different purposes, depending on the target audience. The constant in every pitch is that it ________.
- must be short
- makes a request for support
- includes six elements and about ten slides
- is derived from a corporate narrative
An abbreviated pitch that you memorize is called an elevator pitch. Meant to be given in person, is should be ________.
- brief
- memorized
- connected to a request for more time to make a more complete pitch
- All of these options are correct.
A graphic worksheet that sets forth the value proposition of any startup and allows you to identify stakeholders, and key inputs and outputs is the ________.
- business model generator
- business model canvas
- business mobility canvas
- business meditation and creativity matrix
An agreement between an individual employee and a company that states that innovations belong to the company, even if an individual makes substantial contributions.
- nondisclosure agreement
- work-for-hire agreement
- noncompete clause
- patent
A ________ is a license granting exclusivity, as in the exclusive rights to manufacture and sell a specific product.
- patent
- UI
- nondisclosure agreement
- noncompete clause
Numerical measurements providing data on how people use a product or service results in ________ data.
- qualitative
- quantitative
- lean
- build-measure-learn
More complex data indicating why and how people use or might use a product or innovation is known as ________ data.
- qualitative
- quantitative
- lean
- build-measure-learn
The ________ is the specific request for funding, mentorship, collaborators, and/or business partners that appears in a pitch.
- ask
- request
- pitch
- contest
The student entrepreneurship organization highlighted in this chapter is ________.
- CEO
- CFO
- Future Entrepreneurs of America
- Pitch Competitors International


1.6: The Science of Biology - Branches and Subdisciplines of Biology
- Recognize the various subfields of biology; e.g. microbiology, genetics, evolutionary, etc.
Branches of Biological Study
The scope of biology is broad and therefore contains many branches and subdisciplines. Biologists may pursue one of those subdisciplines and work in a more focused field. The biological branches are divided according to the focus of the discipline and can even be divided based on the types of techniques and tools used to study that specific focus. However, with the increasing amount of basic biological information growing due to advances in technology and databases, there is often cross-discipline and collaboration between branches. For instance, molecular biology and biochemistry study biological processes at the molecular and chemical level, respectively, including interactions among molecules such as DNA, RNA, and proteins, as well as the way they are regulated. Microbiology, the study of microorganisms, is the study of the structure and function of single-celled organisms. It is quite a broad branch itself, and depending on the subject of study, there are also microbial physiologists, ecologists, and geneticists, among others.
Biological Disciplines and Careers
Forensic science is the application of science to answer questions related to the law. Biologists as well as chemists and biochemists can be forensic scientists. Forensic scientists provide scientific evidence for use in courts, and their job involves examining trace materials associated with crimes.Their job activities are primarily related to crimes against people such as murder, rape, and assault. Their work involves analyzing samples such as hair, blood, and other body fluids, including the processing of DNA found in many different environments and materials associated with the crime scenes.
Another field of biological study, neurobiology, is the study of the nervous system, and although it is considered a branch of biology, it is also recognized as an interdisciplinary field of study known as neuroscience. Because of its interdisciplinary nature, this subdiscipline focuses on different functions of the nervous system using molecular, cellular, developmental, medical, and computational approaches.
Additional branches of biology include paleontology, which uses fossils to study life’s history; zoology, which studies animals; and botany, which studies plants. Biologists can also specialize as biotechnologists, ecologists, or physiologists. This is just a small sample of the many fields that biologists can pursue.
Biology is the culmination of the achievements of the natural sciences from their inception to today. Excitingly, it is the cradle of emerging sciences such as the biology of brain activity, genetic engineering of custom organisms, and the biology of evolution that uses the laboratory tools of molecular biology to retrace the earliest stages of life on earth. A scan of news headlines—whether reporting on immunizations, a newly discovered species, sports doping, or a genetically-modified food—demonstrates the way biology is active in and important to our everyday world.
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Key Points
- Biology is broad and focuses on the study of life from various perspectives.
- The branches and subdisciplines of biology, which are highly focused areas, have resulted in the development of careers that are specific to these branches and subdisciplines.
- Branches of biological study include microbiology, physiology, ecology and genetics; subdisciplines within these branches can include: microbial physiology, microbial ecology and microbial genetics.
Key Terms
- genetic engineering : the deliberate modification of the genetic structure of an organism
- forensic : Relating to the use of science and technology in the investigation and establishment of facts or evidence in a court of law.


6.7.1: Prelude to Circuits and DC Instruments
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Electric circuits are commonplace. Some are simple, such as those in flashlights. Others, such as those used in supercomputers, are extremely complex.
This collection of modules takes the topic of electric circuits a step beyond simple circuits. When the circuit is purely resistive, everything in this module applies to both DC and AC. Matters become more complex when capacitance is involved. We do consider what happens when capacitors are connected to DC voltage sources, but the interaction of capacitors and other nonresistive devices with AC is left for a later chapter. Finally, a number of important DC instruments, such as meters that measure voltage and current, are covered in this chapter.


9.3: A Brief Outline of Aegean Clay Tablets
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Chronology and Distribution
At present it remains unclear which of the two Minoan scripts is earlier. The answer to this question lies in the interpretation of the inscriptions from prepalatial seals of the so-called Archanes Script (Godart and Tzedakis 1992: 108, 121–122; Grumach and Sakellarakis 1966; see also Flouda, this volume; Whittaker, this volume): some interpret them as Cretan Hieroglyphic, some as Linear A (Godart 1999; Olivier and Godart 1996; Sakellarakis and Sapouna-Sakellaraki 1997: 326–330). For the present study, however, such a chronological intricacy has no cru- cial relevance, since the earliest preserved clay tablets postdate these Archanes seals. (I do not intend to get involved in a detailed discussion of the absolute chronology of the Aegean Bronze Age here. It suffices to say that Middle Minoan (MM) covers a period of c. 2100 / 2000–1700 / 1600 bc and Late Minoan I (LM I) a period of c. 1700 / 1600–1450 bc; for a detailed discussion, see Warren and Hankey 1989; see also Whittaker, this volume.)
I have already mentioned that Cretan Hieroglyphic tablets are rare: two have been discovered in the Hieroglyphic Deposit of the palace at Knossos, two in the palace of Malia, and one in the palace of Phaistos (Olivier and Godart 1996: 122–123, 172–175, 182–183). Whereas the Phaistos example is of an uncertain date, and the precise date of the Knossian Hieroglyphic Deposit is still a subject of debate (for the latter see the overview of different opinions in Schoep 2001: 147–148), the two tablets from Malia are securely dated to the MM III period (Chapouthier 1930).
A total of some 350 tablets has been discovered amongst Linear A documents, although many in a fragmentary state (Godart and Olivier 1976–1985). The earliest Linear A tablets are an MM IIA example from Knossos (del Freo 2007: 204–205; Schoep [2007] suggests that this may in fact be a Cretan Hieroglyphic tablet) and some 20 tablets from the MM IIB layers of the palace at Phaistos (Pugliese Carratelli 1958). Only a small number of Linear A tablets are of the MM III date while the majority are from the final phase of the administrative use of Linear A, that is, the end of the LM IB period ( c .1450 bc), with the largest number, 147 tablets, coming from Haghia Triada (Ayia Triada) (Halbherr et al. 1977).
At least 5000 tablets with Linear B are known. The earliest are c .650 tablets from the Room of the Chariot Tablets at Knossos (hereafter RCT), dated to LM II or early LM IIIA1 (Driessen 1990: 117). The remainder of the approximately 2800 Knossian tablets are probably of LM IIIA2 date (the most recent overview of disputes concerning the date of the final destruction of Knossos and therefore of the majority of its Linear B tablets, is given in Driessen 2008: 70–72; the corpus of Knossian tablets is published by Chadwick et al . 1986–1998). Other Linear B tablets come from Chania on Crete and major Mycenaean centres on the Greek Mainland. Apart from a few odd examples from the LH IIIA period (Palaima 1983; Shelton 2002–2003), these are all of the LH IIIB date. The Pylos archive with approximately 1100 tablets is the largest in this group (Blegen and Rawson 1966: 92–101). It is followed by nearly 400 tablets from Thebes (Aravantinos et al . 2002; 2008), while there are smaller numbers of tablets from Mycenae, Tiryns and Chania (Andreadaki- Vlasaki and Hallager 2007; Godart 1988; Hallager and Vlasaki 1997; Melena and Olivier 1991; Sacconi 1974).
Shape
There are two principal shapes of Aegean clay tablets: the elongated (also called palm-leaf shaped tablets) and the page-shaped. Linear B makes use of both shapes (the former were used for simple entries, the latter for summarising records, cf. Driessen 1999: 207–208), whereas only page-shaped tablets were employed during the latest stages of Linear A. We do find several elongated tablets inscribed in Linear A, but they are dated to MM II or MM III; no elongated tablets with Linear A have been found amongst the latest surviving, LM IB documents. It should be noted that some LM IB Linear A tablets are far too fragmentary to make a definite decision about their shape, but, since not a single complete elongated tablet has been preserved in any LM IB deposit, we can assume that those tablets that are now fragmentary were also page-shaped when complete. As for Cretan Hieroglyphic tablets, the five preserved examples are all elongated, but thicker than most Linear A and Linear B elongated tablets.
Size
Linear A page-shaped tablets are generally smaller than Linear B ones, as is the average amount of information on them (on complete Linear A tablets the average number of signs per tablet is only 30). The overall proportion of the size of tablets and the ‘crowdedness’ of signs inscribed on them shows that Linear B page-shaped tablets hold a larger amount of information on the available space. As proposed elsewhere (Tomas 2011) there may be several reasons for this: the different nature of the two languages being recorded (Linear A sign-groups, i.e. words, are generally shorter than those of Linear B, see Duhoux 1978: 68), different methods of recording information (for example, Linear A may have used abbreviations more frequently), or different administrative practices (for example, Linear B page-shaped tablets may have been intended to contain more information than Linear A page-shaped tablets; perhaps in Linear A more extensive information was recorded on some other material, possibly perishable, cf. Driessen and Schoep 1999: 392; Olivier 1987: 230).
The small number of Cretan Hieroglyphic and Linear A elongated tablets precludes any sensible comparison to their numerous counterparts in Linear B, especially since the nature of elongated Linear B tablets changed drastically over time. Thus, elongated tablets tend to be smaller and with little text in the RCT, larger in the later Knossian deposits, and much larger and with more abundant text at Pylos. Driessen (1988: 132) notes that very small elongated tablets are frequent in the RCT and completely absent elsewhere: their dimensions are less than 0.6 cm thick, less than 0.2 cm high and less than 0.4 cm long. Their small size suggests that they were meant to be documents that could easily be carried around by individuals. Since it is precisely this type of tablet that in Linear B is most frequently the outcome of cutting (i.e. dividing of a larger tablet into smaller ones) they will be addressed below in more detail.