1 Counting Techniques: Possibility Trees, Multiplication Rule, Permutations

Possibility Trees Winner of set 1 Winner of set 2 Winner of set 3 In a tennis match, the first player to win two sets, wins the game. Question: What is the probability that player A will win the game in 3 sets? Construct possibility tree:

3 Possibility trees and Multiplication Rule  Example: When buying a PC system, you have the choice of  3 models of the basic unit: B1, B2, B3 ;  2 models of keyboard: K1, K2 ;  2 models of printer: P1, P2.  Question: How many distinct systems can be purchased?

4 Possibility trees and Multiplication Rule Select the basic unit Select the keyboard Example(cont.): The possibility tree: Select the printer The number of distinct systems is: 3∙2∙2=12

5 The Multiplication Rule If an operation consists of k steps and  the 1 st step can be performed in n 1 ways,  the 2 nd step can be performed in n 2 ways (regardless of how the 1 st step was performed), ….  the k th step can be performed in n k ways (regardless of how the preceding steps were performed), then the entire operation can be performed in n 1 ∙ n 2 ∙… ∙ n k ways.

6 Multiplication Rule (Example)  Consider the following nested loop: for i:=1 to 5 for j:=1 to 6 [ Statement 1 ; Statement 2. ] next j next i  Question: How many times the statements in the inner loop will be executed?  Solution: 5 ∙ 6 = 30 times (based on the multiplication rule)

Multiplication Rule (Example)  A PIN is a sequence of any 4 digits (repetitions allowed); e.g., 5279, 4346,  Question. How many different PINs are possible?  Solution. Choosing a PIN is a 4-step operation:  Step 1: Choose the 1st symbol ( 10 different ways ).  Step 2: Choose the 2nd symbol ( 10 different ways ).  Step 3: Choose the 3rd symbol ( 10 different ways ).  Step 4: Choose the 4th symbol ( 10 different ways ). Based on the multiplication rule, 10∙10∙10∙10 = 10,000 PINs are possible.

8 Multiplication Rule (Example)  Consider the problem of choosing PINs but now repetitions are not allowed.  Question. How many different PINs are possible?  Solution. Choosing a PIN is a 4-step operation:  Step 1: Choose the 1st symbol ( 10 different ways ).  Step 2: Choose the 2nd symbol ( 9 different ways ).  Step 3: Choose the 3rd symbol ( 8 different ways ).  Step 4: Choose the 4th symbol ( 7 different ways ). Based on the multiplication rule, 10∙9∙8∙7 = 5,040 PINs are possible.

Multiplication Rule and Permutations  Consider the problem of choosing PINs again. Now  a PIN is a sequence of 1, 2, 3, 4 ;  repetitions are not allowed.  Question. How many different PINs are possible?  Solution. Choosing a PIN is a 4-step operation:  Step 1: Choose the 1st symbol (4 different ways ).  Step 2: Choose the 2nd symbol (3 different ways ).  Step 3: Choose the 3rd symbol (2 different ways ).  Step 4: Choose the 4th symbol (1 way ). Based on the multiplication rule, 4∙3∙2∙1 = 4! = 24 PINs are possible.  Note: The number of different PINs in this case is just the number of different orders of 1,2,3,4.

Permutations  A permutation of a set of objects is an ordering of the objects in a row. Example: The permutations of {a,b,c}: abc acb bac bca cab cba  Theorem. For any integer n with n≥1, the number of permutations of a set with n elements is n!.  Proof. Forming a permutation is an n-step operation:  Step 1: Choose the 1 st element ( n different ways ).  Step 2: Choose the 2 nd element ( n-1 different ways ). …  Step n: Choose the n th element (1 way ). Based on the multiplication rule, the number of permutations is n∙(n-1)∙…∙2∙1 = n!

11 Example on Permutations: The Traveling Salesman Problem (TSP)  There are n cities. The salesman  starts his tour from City 1,  visits each of the cities exactly once,  and returns to City 1. Question: How many different tours are possible? Answer: Each tour corresponds to a permutation of the remaining n-1 cities. Thus, the number of different tours is (n-1)!. Note: The actual goal of TSP is to find a minimum-cost tour.