MATH 1150: Freshman Seminar Unit 1. How to count the number of collections The main new problem in this section is we learn how to count the number of ways to pick k objects from a collection of n objects, but without regard to the order in which they were chosen. 1.1. A couple of problems. Suppose we ask ourselves the question: QUESTION 1 : A pizza place has six different toppings, sausage, pepperoni, ham, anchovies, peanut butter, and pineapples. How many different pizzas can we choose if our pizza must have two different toppings? Note that if we were asked to pick two toppings in order, we would be able to easily answer this question. The answer would be 6 5 = 0. But this does not answer the question. After all, a pizza with sausage picked first and then pepperoni is the same as a pizza with pepperoni picked first and then sausage. We ll learn how to answer this question in the next section. QUESTION : How many ways are there for a class with 15 students to form a committee with members? If we had been asked how many ways are there for a class with 15 students to pick a president, vice-president, and secretary, we would know the answer if 15 1 1 = 15!. But this overstates the answer to our question. If we pick the class officers to 1! be Fred as President, Duncan as Vice-President, and Olivia as Secretary, we can use this choice to form a three person committee of Fred, Duncan, and Olivia. But if we had chosen Olivia as President, Fred as Vice-President, and Duncan as Secretary, combining the three officers into a three person committee gives the same committee we got before, i.e., Olivia, Fred, and Duncan. After all, if we are describing a three person committee, the order in which we list the members does not matter. 1.. The Answer to these problems. Let s answer the first question from the last section. For the pizza problem, the trick is to consider the apparently silly question from the previous section, i.e., QUESTION 1 : If a pizza place has six pizza toppings and we are to choose a pizza with two different toppings, how many ways are there to choose the pizza if the order of selection matters? We know that the answer to this question is A = 6 5 = 0, since there are 6 ways to choose the first topping and 5 ways to choose the second topping. We try to solve 1
QUESTION 1, where we count the pizzas without regard to order of selection, by breaking QUESTION 1 into two steps. STEP 1: Pick two different pizza toppings, without regard to order. We will let B be the number of ways to accomplish STEP 1. STEP : Once the two different pizza toppings have been chosen, choose which of the two pizza toppings will be first, and which will be second. Note that it is easy to count C, the number of ways of accomplishing STEP. Indeed, we have two pizza toppings, and we just have to pick one of them to be first. There are ways to do this, so C =. The key point is that A, which is the number of ways to answer QUESTION 1, is the same as the number of ways to accomplish STEP 1 times the number of ways to accomplish STEP. Indeed, for each way of picking two different pizza toppings without regard to order, there are C = ways to accomplish STEP. In other words, we have the equality: A = B C, or by dividing each side by C, B = A C. In this case, A = 0 and C =, so B = 15. This means that there are exactly 15 ways to choose a pizza with two different toppings. Now let s try to answer QUESTION from the last section using the same reasoning. Remember that we want to count the number of person committees we can form from a 15 person class. We start by asking a different question: QUESTION : How many ways are there to select a president, vice-president, and secretary from a 15 person class. If we let A be the answer, note that we know the answer is A = 15 1 1 = 15! 1! As before, we break up the process of selecting the three class officers into two steps: STEP 1: Select a three person committee from a 15 person class. Let B be the number of ways of doing this. B is the answer to QUESTION from the last section.
STEP : From this three person committee, select a president, vice-president, and secretary. Let C be the number of ways of doing this. It is easy to see that C = 1 =! Further, A = B C, since picking a three person committee and then picking president, vice-president, and secretary from this three person committee is the same thing as picking a president, vice-president and secretary from the 15 person class. Dividing each side by C, we obtain: B = A 15 1 1 = = C 1 This gives us the answer to QUESTION. 15! 1!! = 15! 1!!. 1.. General formula. It should not be too hard to see that the idea behind our solutions to QUESTIONS 1 and in the last section are essentially the same. We can generalize these ideas as follows. Theorem 1.1. The number of ways to choose a collection of k objects, without repetition or regard to order of selection, from a collection of n objects is: n! (n k)! k! To make this more concrete, the Theorem asserts that if we want to pick a committee with k people from a group of n people, there are n! ways to do it. Here k and n are any numbers, with k less than n. (n k)! k! To verify this, let A = the number of ways of lining up k people in order from a group of n people. n! From Unit, we know that A = (n k)!. Let B = the number of ways of choosing a committee of k people. Let C = the number of ways of lining up a group of k people in order. From Unit, we know that C = k!. Note that lining up k people from a group of n people in order is the same as first picking k people from a group of n people, and then lining up these k people in order. For each choice of a committee of k people, there are C = k! ways to line them up in order. Using these observations, we see that
A = B C so that B = A n! C = (n k)! k! = n! (n k) k!. The same idea explains why Theorem 1.1 is true. 1.. Application. Example 1.. In a 8 person class, the number of 10 person classes is 8! 18! 10!. To obtain this from Theorem 1.1 from the previous section, we take n = 8, k = 10, so n k = 18. Example 1.. If a pizza place has different toppings, how many ways are there to make a pizza with 1 toppings (and generate much indigestion)? To solve this, let n = and let k = 1, so n k = 9. Then Theorem 1.1 gives the answer:! 9! 1! These fractions with factorials can be annoying to write out. There is a briefer notation. We write n n! n for. The symbol is pronounced n choose k. k (n k)! k! k Using this new notation, we can restate Theorem 1.1 as: Corollary 1.. The number of ways to choose k objects from a collection of n objects n is. k Note that complementary combinations have the same value. For example, 8 = 8! 5! 5! = 8! 8 5!! =. 8 This makes sense in terms of the committee interpretation of. Indeed, the 5 number of ways of choosing a 5 person committee from a group of 8 people is the same as the number of ways of choosing a person committee from the same group. The selection of the 5 people to be on the committee is the same as the selection of the people who are not going to be on the committee. The generalization of this observation is the following fact: Remark 1.5. n n = k (n k)
This ( can ) be verified using a little algebra. By definition, n n! =. Since n (n k) = n n + k = k, we get: ((n k) ) (n (n k))! (n k)! n n! = (n k) k! (n k)! = n! n (n k)! k! =. k To ( illustrate ) this( remark, ) ( note ) the identities: 6 6 =, =. 7 16 ( 1 ) 5 n n Note also that = = 1, which you can verify mathematically using the fact n 0 that 0! = 1. This makes sense because there is only one way to choose a committee of n people from a collection of n people; you have to put everyone on the committee. 1.5. Variants. On a quiz or a test, you are more likely to see a problem that combines the idea of this section with the multiplication principle (Unit ) or the subtraction principle (Unit ). QUESTION 1: A pizza place has meat toppings and 6 vegetable toppings. How many different pizzas can you make by choosing different meat toppings and different vegetable toppings? To answer this, note first that by the multiplication principle, (# of pizza choices) = (# of ways to choose the meat toppings) (# of ways to choose the veggie ( toppings). ) 6 Since there are = 6 ways to choose the meat toppings and = 15 ways to choose the veggie toppings, the answer to our question is: ANSWER: 6 15 = 90. 5 QUESTION : A class has 10 boys and 8 girls, and is picking a co-ed three person committee, where co-ed means that the committee cannot be all boys or all girls. How many committees can be formed? To solve this, use the subtraction principle to note that: (# of co-ed person committees) = (Total # of person committees) MINUS (# of all boy person committees ) MINUS (# of all girl person committees). 18 To count these, note that the total number of person committees is, since the class has 18 people altogether. 10 The number of all boy person committees is
6 and the number of all girl person committees is It follows that our answer is: 18 10 8 ANSWER :. 8. Remark 1.6. Note that in the text we sometimes simplify factorials and sometimes 6 do not. For example, to answer QUESTION 1, we computed = 15 and wrote 6 18 15 in place of. In the answer to QUESTION, we left uncomputed, although we could have computed it. This is mostly an issue of personal taste, with a few caveats. n n = = 1 is a good simplification to make. ( n ) 0 n n = = n is a good simplification to make. ((n 1) ) ( 1) n n n (n 1) = = is usually a good simplification to make. ((n) ) n with big numbers and big differences are rarely useful to simplify. If you try k 0 writing as a number, you will get something quite large and it will take a lot 0 of time. It is better just to leave it as. Later on, we will discuss some ways of 0 understanding an expression like without actually computing it. QUESTION : A basketball team has 1 players, and of the players are centers, of the players are forwards, and 5 of the players are guards. How many starting line-ups can the coach form, if a starting line-up must consist of 1 center, forwards, and guards. To solve this, note that by the multiplication principle, the answer is: (# of ways to choose the center) (# of ways to choose the forwards ) (# of ways to choose the guards). The number of center choices is, since we are picking one center from three 1 choices.
The number of ways to choose the forwards is, since we are picking two forwards from among four of the possible forwards. 5 For the same reason, the number of ways to choose the guards is. It follows that the answer to the question is: ANSWER = 1 5 = 6 10 = 180. 7 QUESTION : How many ways are there to form a letter English word with exactly E s? To answer this, we think of the letter English word by picking two letters to be E, and then the other two letters must be some letter besides E. As a consequence, ANSWER = (# of ways to pick the two letters to be E) (# of ways to decide what the other two letters will be). The # of ways to pick the two letters to be E is, since there are ways to pick two items from a collection of. The # of ways to choose the other two letters is 5, since we ( have ) 5 choices for each of the other two letters. Thus, ANSWER = 5. QUESTION 5: Suppose eight kids are to be divided into two basketball teams, and Jane and Liza are two of the kids. How many ways are there for Jane and Liza to end up on the same team? How many ways are there for Jane and Liza to end up on different teams? Is it more likely than not for Jane and Liza to end up on the same team? Let s try to answer the first part of the question by counting the number of ways for Jane and Liza to end up on the same team. Suppose first that Jane is on the first team and Liza is also on the first team. The number of ways that this can happen is the number of ways to( pick ) two kids from the 6 remaining kids to be on Jane and 6 Liza s team. There are = 15 ways to do this. Now let s count the number of ways ( if) Jane and Liza are on the second team. By the same reasoning as above, there 6 are = 15 ways for this to happen. Thus, we conclude that there are 15+15 = 0 ways for Jane and Liza to wind up on the same team. The easiest way to answer the second part of QUESTION 5 is to use the subtraction principle, which shows that:
8 The # of ways for Jane and Liza to be on different teams = The total # of teams - the # of ways for Jane and Liza to be on the same team. The total # of teams is the number of ways to pick kids to( be) on the first team, 8 since that determines the second team completely. There are = 70 ways to do this. Thus, there are 0 = 70-0 ways for Jane and Liza to wind up on different teams. To answer the last part of QUESTION 5, if the teams are chosen at random, the chances are that Jane and Liza will be on different teams, since there are 0 ways for them to be on different teams, and only 0 ways for them to be on the same team. 1.6. A little probability. In the past when I have taught this course, I have spent a lecture or two on probability. This seems fun because it offers the chance to learn something about odds when you are playing dice or poker, but in practice a number of people end up disliking it, and I suspect nobody has learned enough to make any money. So I ll only cover it briefly. We won t need any serious ideas from probability later in the course, but it may be of some use to understand some simple ideas. Let s go back to the previous section where we discussed the issue of whether Jane and Liza are more likely to be on the same team or not. We concluded that they are more likely not to be on the same team if the teams are chosen randomly. There is a more scientific way to say this. We may say that the probability of Jane and Liza being on the same team is the fraction: ( The # of ways for Jane and Liza to be on the same team )/( The total # of ways to pick the two teams). As we saw in the answer to QUESTION 5 in the last section, the answer is 0 70 =, which is approximately. or percent. 7 For an event to be as likely as not to happen, its probability should be.5 or 50 percent. Definition 1.7. The probability of an event happening is the number of ways for the event to happen divided by the total number of events that can happen.
QUESTION: What is the probability of getting a royal flush, i.e., Ace, King, Queen, Jack, Ten, all of the same suit, from a randomly dealt five card hand from a deck of 5 cards? To answer this question, we need to count the number of ways to deal five cards from a deck of 5 cards and getting Ace, King, Queen, Jack, and Ten, all of the same suit. There are clearly ways to do this, one way for each of the four suits. Next, we count the number of randomly dealt five card hands. There are ( 5 5 ) number of these hands. Using a calculator, you can show that is: ANSWER: 5 =, 598, 960. It follows that are answer 5, 598, 960, which is about 1.5 10 6, or about 1.5 in a million. This means that if you sit in a room and deal yourself 5 cards at random, you should expect to get about one and a half royal flushes every million times you deal the cards. 9 EXERCISES: Explain your answer. (1) Suppose a pizza place has ten meat toppings and five veggie toppings. (a) How many different pizzas can you order with two different meat toppings? (b) How many different pizzas can you order with two different veggie toppings? Are there more pizzas you can order with two meat toppings or two veggie toppings? (c) How many different pizzas can you order with two meat toppings and three veggie toppings? () Suppose a class has 1 kids in it, and five of them are girls and eight are boys. (a) How many ways are there to form a committee with members? (b) How many ways are there to form a committee with members and no girls? (c) How many ways are there to form a committee with members and exactly two boys and two girls? (d) How many ways are there to form a committee with members and at least one boy? () An ice cream shop has thirty-one flavors. (a) How many different sundaes can you make with exactly two different flavors? (b) How many different sundaes can you make with at least three different flavors?
10 () How many six letter English words can be formed with exactly three Z s? (5) Suppose there are n kids, where n is a number and two of them are Liza and Jane. The n kids are to be divided into two teams with n players each. (a) Suppose n = 1 (so there are kids). How many ways are there for Liza and Jane to end up on the same team? (b) Suppose n =. How many ways are there for Liza and Jane to end up on the same team? (c) Suppose n =. How many ways are there for Liza and Jane to end up on the same team? (d) Suppose n = 5. How many ways are there for Liza and Jane to end up on the same team? (e) Compute the probability for Liza and Jane to end up on the same team in the cases n = 1,,, and 5. Do you see a pattern? If n = 6, can you predict the probability for Liza and Jane to end up on the same team? In this case, do Liza and Jane have at least an even chance of being on the same team? (6) What is the probability of drawing a straight flush from a randomly dealt five card hand? (a straight flush means five consecutive cards of the same suit. For these purposes, Ace is one higher than King, but Ace and Two are not consecutive). (7) Suppose there are ten kids, and two of them are Liza and Jane. Suppose eight of them are chosen to play basketball, and those eight are divided into two teams. How many ways are there to do this so that Liza and Jane end up on the same team? If this is done, what is the probability of Liza and Jane ending up on the same team?