# CIS 2033 Lecture 5, Fall

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1 CIS 2033 Lecture 5, Fall Instructor: David Dobor September 13, Supplemental reading from Dekking s textbook: Chapter2, 3. We mentioned at the beginning of this class that calculus was a prerequisite for this course. Counting, however, is not. So we consider the very basics of counting. Introduction A basketball coach has 20 players available. Out of them, he needs to choose five for the starting lineup, and seven who would be sitting on the bench. In how many ways can the coach choose these 5 plus 7 players? It is certainly a huge number, but what exactly is it? In this lecture, we will learn how to answer questions of this kind. More abstractly, we will develop methods for counting the number of elements of a given set which is described in some implicit way. Now, why do we care? The reason is that in many models, the calculation of probabilities reduces to counting. Counting the number of elements of various sets. Suppose that we have a probability model in which the sample space, Ω, is finite, and consists of n equally likely elements. So each element has probability 1/n. Suppose now that we re interested in the probability of a certain set, A, which has k elements. Since each one of the elements of A has probability 1/n, and since A has k distinct elements, then by the additivity axiom, the probability of A is equal to k 1/n. Therefore to find the probability of A, all we have to do is to count the number of elements of Ω and the number of elements of A, and so determine the numbers k and n. Of course, if a set is described explicitly through a list of its elements, then counting is trivial. But when a set is given through some abstract description, as in our basketball team example, counting can be a challenge.

4 cis 2033 lecture 5, fall first stage, there are only 25 available letters that can be used. We only have 25 choices at the second stage. Now, let us start dealing with the digits. We choose the first digit, and we have 10 choices for it. However, when we go and choose the next digit we will only have 9 choices, because 1 of the digits has already been used. At this point, 2 digits have been used, which means that at the last stage we have only 8 digits to choose from. So by multiplying these numbers, we can find out the answer to this question, the number of license plates if repetition is prohibited. Number of different license plates = ( repetitions not allowed ) Example 3. Now suppose that we start with a set that consists of n elements. What we want to do is to take these n elements and order them. A terminology that s often used here is that we want to form a permutation of these n elements. One way of visualizing permutations is to say that we re going to take these elements of the set, which are unordered, and we re going to place them in a sequence of slots. So we create n slots. And we want to put each one of these elements into one of these slots. How do we go about it? We think of putting the elements into slots, one slot at a time. We first consider the first slot. We pick one of the elements and put it there. How many choices do we have at this stage? We have n choices, because we can pick any of the available elements and place it in that slot. Next, we pick another element and put it inside the second slot. How many choices do we have at this step? Well, we have already used one of the available elements, which means that there s n 1 elements to choose from at the next stage. At this point, we have used 2 of the elements. There are still n 2 elements left. We pick one of these n 2 elements and put it in the third slot this point. We continue this way. At some point we have placed n 1 of the elements into slots. There s only one element left, and that element, necessarily, will get into the last slot. There are no choices to be made at this point. So the overall number of ways that we can carry out this process, put the elements into the n slots, by the counting principle is going to be the product of the number of choices that we had at each one of the stages. So it s this product: Figure 2: We can think of permutations of n elements as a number of different ways of placing unordered elements into a sequence, here represented by slots into which we place these elements. n (n 1) (n 2)

5 cis 2033 lecture 5, fall And this product we denote as a shorthand this way, n! which we read as n factorial. n! is the product of all integers from 1 all the way up to n. And in particular, the number of permutations of n elements is equal to n!. Example 4. Let us now consider another example. We start again with a general set, which consists of n elements. And we re interested in constructing a subset of that set. In how many different ways can we do that? How many different subsets are there? Let us think of a sequential process through which we can choose the subset. The sequential process proceeds by considering each one of the elements of our set, one at a time. We first consider the first element, and here we have 2 choices. Do we put it inside the set or not? So 2 choices for the first element. Then we consider the second element. Again, we have 2 choices. Do we put it in the subset or not? We continue this way until we consider all the elements. There s n of them. And the overall number of choices that we have is the product of , this product taken n times (because we make that choice of either including or excluding an element for each of the n elements. This of course works out to be 2 n. At this point, we can also do a sanity check to make sure that our answer is correct. Let us consider the simple and special case where n is equal to 1, which means we re starting with a set with 1 element and we want to find the number of subsets that it has. According to the answer that we derived, this should have 2 1, that is 2 subsets. Which ones are they? One subset of this set is the set itself and the other subset is the empty set. So we do have, indeed, 2 subsets out of that set, which agrees with the answer that we found. Notice that when we count subsets of a given set, we count both the set itself, the whole set, and we also count the empty set.

6 cis 2033 lecture 5, fall Another Example We will now use counting to solve a simple probabilistic problem. We have in our hands an ordinary sixsided die which we are going to roll six times - this is our probabilistic experiment. We re interested in the probability of the event that the six rolls result in different numbers. So let us give a name to that event and call it event A: We wish to calculate the P(A). But before we can even get started answering this question, we need a probabilistic model. We need to make some assumptions, and the assumption that we re going to make is that all outcomes of this experiment are equally likely. This is going to place us within a discrete uniform probabilistic model so that we can calculate probabilities by counting. In particular, as we discussed earlier in this lecture, the probability of an event A is going to be the number of elements of the set A, the number of outcomes that make event A occur, divided by the total number of possible outcomes, which is the number of elements in our sample space P(A) = Number of Elements in A Total Number of Possible Outcomes So let us start with the denominator, and let us look at the typical outcomes of this experiment. A typical outcome is something like this sequence, 2, 3, 4, 3, 6, 2. That s one possible outcome How many outcomes of this kind are there? Well, we have 6 choices for the result of the first roll, 6 choices for the result of the second roll, and so on. And since we have a total of 6 rolls, this means that there is a total of 6 to the 6th power possible outcomes, according to the Counting Principle. And by the way, since we have so many possible outcomes and we assume that they are equally likely, the probability of each one of them would be 1 over 6 6 P( Sequence {2, 3, 4, 3, 6, 2 } occurs ) = ( Incidentally, that s the same number you would get if you were to assume, instead of assuming directly that all outcomes are equally likely, to just assume that the different rolls are rolls of a fair sixsided die, so the probability of getting a 2, say, is 1/6, and also that the different rolls are independent of each other. So in that case, the probability, let s say, of the particular sequence 2, 3, 4, 3, 6, 2 would be the probability of obtaining a 2, which is 1/6, times the probability that we get a 3 at the next roll, which is 1/6, times 1/6 times 1/6 and so on, and we get the same answer, 1 over 6 6. So we see that this assumption of all outcomes being equally

7 cis 2033 lecture 5, fall likely has an alternative interpretation in terms of having a fair die which is rolled independently 6 times.) Now, let us look at the event of interest, A. What is a typical element of A? A typical element of A is a sequence of 6 rolls in which no number gets repeated. So, for example, it could be a sequence of results of this kind: 2, 3, 4, 1, 6, 5. So all the numbers appear exactly once in this sequence. So to compute the number of outcomes that make event A happen, we basically need to count the number of permutations of the numbers 1 up to 6, so these 6 numbers have to appear in an arbitrary order. In how many ways can we order 6 elements? As discussed earlier, this is equal to 6!. So we have now the probability of event A: P(A) = 6! 6 6 Combinations Let us now study a very important counting problem, the problem of counting combinations. What is a combination? We start with a set of n elements. And we re also given a non-negative integer k. We want to construct or to choose a subset of the original set that has exactly k elements. In different language, we want to pick a combination of k elements of the original set. In how many ways can this be done? Let us introduce some notation. We use the notation ( n k ), which we read as "n-choose-k," to denote exactly the quantity that we want to calculate, namely the number of subsets of a given n-element set, where we only count those subsets that have exactly k elements. Figure 3: We can think of combinations as number of different ways in which you we can choose a subset of size k from a set of size n. How are we going to calculate this quantity? Instead of proceeding directly, we re going to consider a somewhat different counting problem which we re going to approach in two different ways, get two different answers, compare those answers, and by comparing them, get an equation which is going to give us the desired answer.

8 cis 2033 lecture 5, fall We start, as before, with our given set that consists of n elements. But instead of picking a subset, what we want to do is to construct a list, an ordered sequence, that consists of k distinct elements taken out of the original set. So we think of having k different slots, and we want to fill each one of those slots with one of the elements of the original set. In how many ways can this be done? Well, we want to use the counting principle, so we want to decompose this problem into stages. We choose each one of the k items that go into this list one at a time. We first choose an item that goes to the first position, to the first slot. Having used one of the items in that set, we re left with n 1 choices for the item that can go into the second slot. And we continue similarly. When we re ready to fill the last slot, we have already used k 1 of the items, which means that the number of choices that we re going to have at that stage is n k + 1. Thus, the number of ways to fill these k slots is n (n 1)... (n k + 1) Figure 4: We fill the list of k slots with elements from a set of size n, one at time. Any of the n items can go into the first position in the list, then any of the n 1 remaining items can go into the second position, and so on, all the way down to the k-th position in the list, where any of the remaining (n - k + 1) remaining items can go. We then simply rewrite this product as n! (n k)!. At this point, it s also useful to simplify that expression a bit. We n! observe that this is the same as. (Why is this the case? You can (n k)! verify this is correct by moving the denominator to the other side. When you do that you realize that you have the product of all terms from n down to n k + 1. And then you have the product of n k going all the way down to 1. And that s exactly the product, which is the same as n!.) Number of different ways to fill the list of k elements = = n (n 1)... (n k + 1) = n! (n k)! So this was the first method of constructing the list that we wanted. How about a second method? What we can do is to first choose k items out of the original set, and then take those k items and order them in a sequence to obtain an ordered list. We construct our ordered list in two stages. In the first stage, how many choices do we have? that s the number of subsets with k elements out of the original set. This number, we don t know what it is - that s what we re trying to calculate. But we have a symbol for it: It s ( n k ). Figure 5: The second way of building How about the second stage? We have k elements, and we want to arrange them in a sequence. That is, we want to form a permutation the list is shown in blue. We first choose k items out of the original set - we don t yet know what that number is, but we have a symbol for it: ( n k ) - and then take those k items and order them in a sequence to obtain an ordered list.

10 cis 2033 lecture 5, fall Let s look at another extreme case now, the coefficient ( n 0 ). This time let us start from the formula. Using the convention that we have, this is equal to 1: = n! 0 0! n! = 1 Is it the correct answer? How many subsets of a given set are there that have exactly zero elements? Well, there s only one subset that has exactly 0 elements, and this is the empty set,. Now, let us use our understanding of those coefficients to solve a somewhat harder problem. Suppose that for some reason, you want to calculate this sum: n k=1 k How do we compute this sum? One way would be to use the formula for these individual terms in the sum, do a lot of algebra. And if you re really patient and careful, eventually you should be able to get the right answer. But this would be very painful! Let us think whether there s a clever way, a shortcut, of obtaining this answer. Let us try to think what this sum is all about. This sum includes the term, ( n 0 ), which is the number of zero-element subsets. The sum also includes the term ( n 1 ), which is the number of subsets that have one element. And we keep going all the way to the number of subsets that have exactly n elements. So just by rewriting the sum, we have: n k=1 = k = n = Number of all subsets of a set of n elements. So we re counting zero-element subsets, one-element subsets, all the way up to n element subsets. But this is really the number of all subsets of our given set, and this is a number that we already know! It is the number of subsets of a given set with n elements and is equal to 2 n : n k=1 = 2 n k So by thinking carefully and interpreting the terms in this sum, we were able to solve this problem very fast, something that would

17 second approach is a little easier, because we never had to involve any p s in our calculation. cis 2033 lecture 5, fall

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