Math 221 Integrated Learning System Week 2, Lecture 1 Basic Counting Principles; Permutations and Combinations
Venn Diagrams S A B Venn diagrams are one of the basic ways of counting objects in a set when the set has a mixture of characteristics with some objects having only one of the characteristics and others having more than one of the characteristics. What do we mean by characteristic? A characteristic may be any qualitative or quantitative attribute that characterizes an object, such as color, ethnicity, gender, behavior, etc. In the examples to come later, the concept of characteristic should become more clear. For now let us say that in the Venn diagram above we have a population of size S that is represented by the area inside the rectangle. Within this population there are two characteristics A and B that are attributes of some of the objects in S. The areas inside the circles represent those objects in S that have characteristics A or B. The overlap of the circles indicates that some of the objects in S have both characteristic A and characteristic B. To see how this can be suppose that objects with characteristic A eat breakfast every day and those with characteristic B eat lunch every day. Then those with both characteristics would eat both breakfast and lunch everyday. Those outside of the circles would eat neither breakfast nor lunch every day. The symbol AB is read A intersect B, or A and B, and is used to indicate those objects in sets A and B that have the characteristics making them members of both sets.
S A B Let A represent the number of objects that have the characteristic making them a member of set A. Let B represent the number of objects that have the characteristic making them a member of set B. Then AB represents the number of objects that have both the characteristic making them a member of set A and the characteristic making them a member of set B. If we want to know the number of objects in sets A and B combined, we cannot just add A to B because AB is a part of both sets and would be counted twice. So we must subtract AB from the sum A + B avoid this double counting. The result is AB = A + B - AB. We call this A union B, or A or B.
S AB’ A’B A’B’ The objects that are in set A but not in set B we denote as AB’. B’ consists of the members of set S that do not have the characteristic making them a member of set B, and is called the complement of set B. This is read as A intersect B not. Similarly, the objects that are in set B but not in set A are denoted as A’B. And the objects that are in set S but in neither set A nor set B are denoted as A’B’.
Example S A B Before you read on take a minute an see if you can figure out the solutions to these questions. Think about how to use the formula AB = A + B - AB to find the solutions. Then scroll down to see the solutions. 1. AB = 75 + 50 – 20 = 105 2. What is being asked here is what is in the green circle but not in the red circle. Or in other words, what objects have only characteristic A? We know that 75 objects have characteristic A and 20 objects share both characteristics. So the number that have only characteristic A must be 75 – 20 or 55. 3. What is being asked here is how many objects have neither characteristic A nor characteristic B. This is the complement of question 1. So the solution is 150 – 105 or 45. 4. The solution to this question is derived similarly to the solution to question 2. How many objects have characteristic B but not characteristic A? 50 – 20 = 30. The techniques shown can be extended to three or more characteristics. Think about how that could be done. S contains 150 objects. A contains 75 objects; B contains 50 objects and AB contains 20 objects. 1. How many objects are in AB? 2. How many objects are in AB’? 3. How many objects are in A’B’? 4. How many objects are in A’B?
Multiplication Principle Lets say that you are preparing to cook a meal. You want to serve a meat dish, a starch dish, a vegetable dish and a dessert. You have two different meats in your refrigerator, Three different frozen vegetables, two choices of starches and two different desserts. How many different meals are possible? Go on to the next slide The solution to this problem is an example of the multiplication principle.
Let’s see how this looks in what we call a decision tree. Let’s say that first you choose a meat. And then you choose a vegetable. Next you choose a starch. An finally, you choose a dessert. You have two choices at step 1, three choices at step 2, two choices at step 3, and two choices at step 4. Go on to the next slide Let’s see how this looks in what we call a decision tree.
There are 2 times 3 times 2 times 2 equals 24 different meals you could prepare from this collection of foods. This, in general, is the multiplication principle. If you can break down a process into a sequence of steps with a quantifiable number of possible choices or outcomes at each step, then the total number of possible outcomes of the process is the number of possible outcomes of the first step times the possible number of outcomes at the second step times the possible number of outcomes at the third step … times the possible number of outcomes at the last step. Many processes can be modeled in this manner.
Example How many different sequences of five cards can be dealt if you are dealing from a standard deck of 52 cards? There are 52 ways to deal the first card, 51 ways to deal the second card, 50 ways to deal the third card, 49 ways to deal the fourth card and 48 ways to deal the fifth card. Go on to the next slide. Note: This is not the number of possible five card poker hands that can be dealt from a standard deck of 52 cards. In poker hands, the sequence in which the cards are dealt has no effect on the value of the hand.
Factorials n! (read n factorial) represents the product of the first n natural numbers. It is calculated in this manner: n! = n(n - 1)(n - 2)...(2)(1) Examples: Factorials are used extensively in counting as we will see from the next few slides. Factorials build up very rapidly. Your TI-83 calculator can only calculate up to 69!. Try it yourself. To calculate 69!, key in 69 then press the MATH key. Use the right tab key to move the highlighted menu item to PRB. Then use the down tab key to move the highlighted item to item 4. Press the enter key twice. You should have the following displayed on your screen: 1.711223524E98. The E98 at the end of this number means that the number is multiplied by 10 to the 98th power. Now try to calculate 70!. You should get the following message on your screen: ERR:OVERFLOW 1:Quit 2:Goto This means that the calculator could not calculate the number you requested. Just press the enter key and the message will go away. NOTE: 0!=1 by definition
Permutations A permutation of a set of objects is an arrangement of the objects in a specific order without repetition.. Example: How many ways can the five letters a through e be arranged without repetition? (We don’t care whether or not the arrangements spell a word.) There are 5 ways to choose the first letter, 4 ways to choose the second letter, 3 ways to choose the third letter 2 ways to choose the fourth letter and 1 way to choose the fifth letter. Notice that the calculation of the number of permutations of n objects when we are using all of the objects is the same as the calculation of n!. This is more interesting when we are using only some of the n objects. (Note: A lower case italic n will be used to indicate an integer valued number of objects at all times in this course.)
Now, suppose we only want to use three of the five letters a - e at a time. How many permutations would we now have? This gives us a clue as to how to what a general formula for the calculation might be. Think about what the general formula could be before you go on to the next slide.
Next we will look at some examples.
Examples The last two are for you to do on your calculator. There are two ways to do this on your calculator. You can use the factorial function on the calculator and enter the whole formula or you can use the permutations function on your calculator (if you have a TI-83). To use the permutations function on your calculator do the following: Enter 37 then press the math key. Tab over to PRB. Tab down to nPr. Press the enter key. Enter 9 and press the enter key again. You should have the answer 4.51435863E13. This is a very large number. Try this yourself on the fourth example. You should get 39070080. What if we are not interested in how many different arrangements we can make of n objects taken r at a time, but are only interested in how many different sets of r objects we can make from n objects? This is a little different problem. Permutations counts the number of different sets along with the number of different arrangement of objects in each set. If we are not interested in the arrangements we are interested in what we call combinations, not in permutations. Go on to the next slide.
Combinations A combination of n objects taken r at a time without repetition is a subset of size r of the n objects. The arrangement of the objects does not matter.
We know that the number of permutations of n objects taken r at a time is Go on to the next slide But this number takes into account all of the possible arrangements of the r objects. In combinations the different arrangements are not counted as unique. How can we sort of “uncount” the arrangements?
One way to accomplish this is to divide the number of permutations by the number of arrangements. Remember that the number of arrangements of r objects is r!. Hence the number of combinations of n objects taken r at a time is: I prefer to use the notation listed last. This is usually read as n choose r.
Examples Now we can answer questions like, what is the number of possible 5 card poker hands that can be dealt from a standard deck of 52 cards? This is the same as, how many combinations of 52 objects taken 5 at a time are there? Note that this is a much smaller answer than the number of ways a five card hand can be dealt from a standard deck of 52 cards. Remember that answer was 311,875,200.
Example How many possible 7 card hands that have exactly 3 clubs and 4 spades? Before you go on to the next slide to view the solution, think about how you would solve this problem using what we have covered in this lecture. This is not a simple combination problem.
Notice that to get the final solution we had to use the multiplication principle!
Example for You A catering service offers 8 appetizers, 10 main courses,and 7 desserts. A banquet committee is to select 3 appetizers, 4 main courses, and 2 desserts. How many ways can this be done? Share your solution to this problem with the other students in the TDA.
Example for You An electronics store receives a shipment of 30 graphing calculators, including 6 that are defective. Four of these calculators will be sent to a local high school. (A) How many selections can be made? (B) How many of these selections will contain no defective calculators? Share your solution to this problem with the other students in the TDA.