Topic 3: Counting Techniques CEE 11 Spring 2002 Dr. Amelia Regan These notes draw liberally from the class text, Probability and Statistics for Engineering and the Sciences by Jay L. Devore, Duxbury 1995 (4th edition)

Counting Techniques n When the various outcomes of an experiment are equally likely then the task of computing probabilities reduces to counting. In particular, if N is the number of outcomes in the sample space and N(A) is the number of outcomes contained in an event A, then n The product rule for ordered pairs: If the first element or object of an ordered pair can be selected in n 1 ways and for each of these n 1 ways the second element of the pair can be selected in n 2 ways, then the number of pairs is n 1 n 2

Review Questions n Problem #57, chapter 2 u For any two events A and B with P(B) > 0 Show that P(A|B) + P(A’|B) = 1.0 Do this mathematically or with a venn diagram

Review Question n Solution n Note -- we use the law of total probability

Review Questions n Problem #58, chapter 2 u Show that for any three events A, B and C with P(C) > 0 F (AUB|C) = P(A|C)+P(B|C)-P(A intersection B|C)

Counting Techniques n For small problems, we sometimes construct a tree diagram to enumerate the different possible combinations. Assume that a UCI student must choose from three dorms and that he or she must select a room with zero, one or two roommates. The tree diagram below shows the nine possible combinations. D1D1 D2D2 D3D3 r0r1r2r0r1r2 r0r1r2r0r1r2 r0r1r2r0r1r2

Permutations n The number of permutations of n objects is equal to n(n-1)(n-2)…(2)(1) or n! For example, if a traveling salesman must visit 10 customers on a particular day there are 10! different routes that he may take. n Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from the n objects is denoted by P k,n n The number of permutations of size k that can be constructed from n objects is equal to n(n-1)(n-2)…(n-k+1)

Permutations n Sometimes we need to determine the number of permutations possible when some of the objects in a group are indistinguishable from each other. For example: (Ross, 1988, p5) How many different letter arrangements can be formed using the letters PEPPER?

Permutations Solution: First note that there are 6! Permutations of the letters P 1 E 1 P 2 P 3 E 2 R when the 3 P’s and 2 E’s are distinguished from each other. However consider any one of these permutations -- for instance, P 1 P 2 E 1 P 3 E 2 R. If we now permute the P’s among themselves and the E’s among themselves then the resultant arrangement would still be of the form PPEPER. That is, all 3!2! Permutations are of the form PPEPER. Hence there are 6!/3!2! = 60 possible letter arrangements of the letters PEPPER.

combinations n We are often interested in determining the number of different groups of k objects may be formed from a group of n objects n Given a set of n distinct objects, any unordered subset of size k of the objects is called a combination. The number of combinations of size k that can be formed from n distinct objects is denoted by or sometimes by C k,n

Class exercise n To successfully route luggage from one airport to another the destination airport is identified by three letters. For example, Los Angeles is LAX, Orange County is SNA. How many airports can be served by this identification system?

Class exercise n A student looking for a good used car finds ten cars in his price range advertised on the web. Since he has to ride the bus to look at the cars in person he has only enough time to look at four of them. n Order matters in this case because if he finds a car he likes at any time in the process he’ll buy it. In how many ways can the four cars be selected from the set of ten? n If now we assume that he’ll look at all four cars and then make his decision (order does not matter). In how many ways can the four cars be selected from the set of ten?

Class exercise n Using the principle that n What is the probability of dealing a perfect bridge hand? (13 cards of the same suit?)