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Multiplication Rule. A tree structure is a useful tool for keeping systematic track of all possibilities in situations in which events happen in order. The next example shows how to use such a structure to count the number of different outcomes of a tournament. Tournament Play .
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A tree structure is a useful tool for keeping systematic track of all possibilities in situations in which events happen in order. The next example shows how to use such a structure to count the number of different outcomes of a tournament.
Tournament Play Teams A and B are to play each other repeatedly until one wins two games in a row or a total of three games. a. How many ways can the tournament be played? b. Assuming that all the ways of playing the tournament are equally likely, what is the probability that five games are needed to determine the tournament winner?
The possible ways for the tournament to be played are represented by the distinct paths from “root” (the start) to “leaf” (a terminal point) in the tree shown Why should we stop the tree after 5 games?
The Multiplication Rule Suppose a person has four different pants (A, B, C,and D) and three different shirts (X, Y,and Z). In how many different ways the person can the person dress up?
A X Start B Y C D Z
This process is a sequence of events First Event: Choose the pants Second Event: After you choose the pant choose the shirt.
Notice that each pant can be chosen in four different ways. After a pant is chosen a shirt (to match up) can be chosen in three different ways. Total Number of Ways (as a 2-word or ordered pair) 4 3 ______ ________ Size First Event Size Second Event
EXAMPLES A typical PIN (personal identification number) is a sequence of any four symbols chosen from the 26 letters in the alphabet and the ten digits, with repetition allowed. How many different PINs are possible? Solution: Typical PINs are CARE, 3387, B32B, and so forth. You can think of forming a PIN as a four-step operation to fill in each of the four symbols in sequence.
Step 1: Choose the first symbol. Step 2: Choose the second symbol. Step 3: Choose the third symbol. Step 4: Choose the fourth symbol. 36363636 By the multiplication rule, there are 36 36 36 36 = 364 = 1,679,616 PINs
Number of PINs without Repetition Now form PINs using four symbols, either letters of the alphabet or digits, and suppose that repetition is not allowed. a. How many different PINs are there? b. If all PINs are equally likely, what is the probability that a PIN chosen at random contains no repeated symbol?
There are 36 ways to choose the first symbol,35 ways to choose the second (since the first symbol cannot be used again), 34 ways to choose the third (since the first two symbols cannot be reused), and 33 ways to choose the fourth (since the first three symbols cannot be reused). 36353433 The multiplication rule can be applied to conclude that there are 36 35 34 33 = 1,413,720 different PINs with no repeated symbol.
Thus the probability that a PIN chosen at random contains no repeated symbol is In other words, approximately 84% of PINs have no repeated symbol.
Strings or Words If S is a nonempty finite set of characters, a string over S is a finite sequence of elements of S. The number of characters in a string is called the length of the string. The null string over Sis the “string” with no characters. There is one string of length zero. Let A={a, b, c} • How many 3-strings can be made where the characters are elements from A? • How about the number of 5-strings?
r-Permutations Let A={a, b, c} A 2-permutation of A is an ordering of any two distinct elements of A in a row (2-strings no repetitions) ab ac ba bc ca cb How to count them? _________ __________ 1stcharacter 2ndCharacter P(3,2)=No. of 2-permutations of a set with 3 elements.
Let A={a, b, c} A 1-permutation of A is an ordering of one elements of A in a row (1-strings) a b c How to count them? _______ 1st Character P(3,1)=No. of 1-permutations of a set with 3 elements.
A 3-permutation of A is an ordering of any three distinct elements of A in a row (3-strings no repetitions). There are six permutations of A abc acb cba bac bca cab How to count them? ________ _________ _________ 1st 2nd 3rd P(3,3)= No. of 3-permutations of a set with 3-elements
Calculate P(3,4) P(3,0) NOTE: P(3,3) is simply denoted P(3) and represents the permutations of a 3-set. DEFINITION: n a positive integer 3!= 0! = 1 by definition
Formula for P(n,r) Let n be a positive integer and r a non-negative integer less than or equal to n