combinaTorial Analysis Principle of counting Permutations Combinations

Principal of counting If some event can occur in 𝑛 1 different ways, then a second event can happen in 𝑛 2 different ways then a third event can happen in 𝑛 3 different ways … Then the total number of possible outcomes of all events is 𝑛 1 * 𝑛 2 * 𝑛 3 * … Example how many license plates can there be with 2 letters followed by 3 digits where the first digit is not a 0? 26 * 26 * 9 * 10 * 10 = 608,400

Factorial Notation The product of all positive integers from 1 to n inclusively is denoted as n! n! = 1 * 2 * …* (n-1) * n Example) What is 4! ? 1*2*3*4 = 24 Notice n! = n * (n-1)! So we define 0! = 1 so that 1! = 1*0!

Examples of factorials How many ways can you arrange the three letters ABC? S = {ABC, ACB, BAC, BCA, CAB, CBA} = 6 = 3! How many ways can we choose the first letter? 3 Once that first letter is chosen, how many ways can we choose the second letter? 2 Once the first and second letters are chosen, now many way can we choose the third letter? 1 Total number if choices? 3*2*1 = 6

Example) Choose 2 from 5 Given the set of letters ABCDE, how many ways can you choose 2 letters where the order of the letters doesn’t matter? Well, there are 5! = 120 permutations of these letters, but we are only interested in the first 2 letters, so in the 120 permutations, that would include the permutations ABCDE, ABCED, ABDCE, ABDEC, ABECD, ABEDC BACDE, BACED, BADCE, BADEC, BAECD, BAEDC The 2 letters A and B show up as the first 2 letters 12 times, in the 120 permutations.

Example) 5 choose 2 ABCDE, ABCED, ABDCE, ABDEC, ABECD, ABEDC BACDE, BACED, BADCE, BADEC, BAECD, BAEDC There are 2! ways to list A and B and there are also there 3! ways to arrange the remaining letters CDE. So, if we were to count the 5! = 120 permutations of ABCDE, we will have to divide them by all the number of sets of duplicates. 𝑃𝑒𝑟𝑚(𝐴,𝐵,𝐶,𝐷,𝐸) (𝑃𝑒𝑟𝑚 𝐴,𝐵 ∗𝑃𝑒𝑟𝑚(𝐶,𝐷,𝐸)) = 𝑛! 𝑟 !∗ 𝑛−𝑟 ! Where r is the number of items you are choosing from n

Binomial Coefficients 𝑛 𝑟 = 𝑛! 𝑟 !∗ 𝑛−𝑟 ! = 𝑛∗ 𝑛−1 ∗…∗(𝑛−𝑟+1) 1∗2∗…∗𝑟 Example) How many ways can you choose 3 unique letters from the alphabet? 26 3 = 26! 3!∗23! = 26∗25∗24 1∗2∗3 =2600

Ordered Partitions Example) Given the letters of the alphabet, how many ways can you house 3 unique letters as your first set, then from the remaining letters pick another 2 unique letters? 26 3 ∗ 23 2 =657800

Ex) Win, Place, Show (Horse racing) There are 16 horses in a race, each having an equal chance to win. What are your chances to win? Win) Your horse must come in 1st place. 1 16 Place) Your horse must come in 1st or 2nd ? 2 16 Show) Your horse must come in 1st , 2nd or 3rd ? 3 16

Ex) Exacta (Horse racing bet) How many possible outcomes are there for these bets? Exacta) You must pick the 1st AND the 2nd place horse 16∗15=240 Exacta Box) You pick 2 horses and those 2 horses must come in 1st and 2nd in any order 16∗15 1∗2 =120

Ex) Trifecta (Horse racing bet) How many possible outcomes are there for these bets? You must pick the 1st AND 2nd AND 3rd place horse 16∗15∗14=3360 Trifecta Box) You pick 3 horses and those 3 horses must come in 1st , 2nd and 3rd in any order 16∗15∗14 1∗2∗3 =560

Ex) Super Bowl Football has 32 teams divided into 2 sets called conferences. The two conferences are disjoint sets of 16 teams. Each conference produces a winner and those 2 winners play each other in the Super Bowl. How many different combinations of teams may play in the super bowl? 16∗16=256 Not 32 2 because they are disjoint sets of 16 teams

Ex) Super Bowl continued Each conference has it’s own championship game prior to the Super Bowl. How many different combinations of teams may play in the two conference championship games? 16∗15 1∗2 ∗ 16∗15 1∗2 =14,400