Combinations, Permutations, and the Fundamental Counting Principle

 If one event can happen m ways and another event can happen n ways, the two events can happen in sequence m x n ways.  You want to know how many different ice cream sundaes you can make with the following criteria: ◦ 1 flavor of ice cream (selected from 6 flavors) ◦ 1 flavor of syrup (selected from 3 flavors) ◦ 1 type of topping (selected from 8 choices)

 6 * 3 * 8 =  144  There are 144 different ice cream sundaes possible, choosing one item from each category.  If you wanted to choose more than one off a list, you would need to use “combinations”.

 Factorial is denoted by !  This represents the number of different ordered arrangements of n distinct objects (n!)  You multiply the numbers in descending order until you reach 1: ◦ Example: 5! = 5 * 4 * 3 * 2 * 1 = 120 ◦ Example: 3! = 3 * 2 * 1 = 6 ◦ Rule for factorials: 0! = 1 (Special Case)

 How many ways items can be selected from a larger group  Order does not matter  n C r :n = total number in data set r = number of items taken at a time r ≤ n r, n ≥ 0, must be integers  n C r = n! / (n-r)!r!

 Example: Use combinations for items on a pizza (order of pepperoni, sausage, etc. doesn’t matter).  How many different 3 topping pizzas can be made if there are 12 different toppings to choose from?  12 C 3 = 12! / (12-3)! 3! = 12 * 11 * 10 * 9! / 9! * 3 * 2 * 1 = 12 * 11 * 10 / 3 * 2 * 1 = 220 There are 220 different 3 topping pizzas possible. **Note about 9!

 6 types of ice cream (I want 2 different ones)  3 types of syrup (I still only want 1 kind)  8 types of toppings (I want 3 different ones)  6 C 2 * 3 C 1 * 8 C 3 =  6 C 2 = 6! / (6-2)! 2! = 6*5*4! / 4!*2*1 = 15  3 C 1 = 3! / (3-1)! 1! = 3*2! / 2!*1 = 3  8 C 3 = 8! / (8-3)! 3! = 8*7*6*5! / 5!*3*2*1 = 56  So 15 * 3 * 56 = 2520  There are 2,520 different sundaes possible with the above criteria.

 How many ways items can be chosen from a larger group  Order does matter  n P r = n! / (n-r)!  There will be a larger number of permutations than combinations.  Combination locks are actually permutation locks.

 How many ways can 10 people running a race finish 1 st, 2 nd, and 3 rd ?  10 P 3 = 10! / (10-3)! = 10 * 9 * 8 * 7! / 7! = 10 * 9 * 8 = 720  There are 720 ways 10 people running a race can finish 1 st, 2 nd, and 3 rd.

 If the same number or letter is in a sequence, the sequence may look the same even if you have rearranged the digits/letters. To determine how many distinguishable permutations there are:  n! / n1!*n2!*n3!... where the denominator terms are the repeats, such as this…  MISSISSIPPI: 11! / 4! * 4! * 2! = ◦ 4! = 4 I’s ◦ 4! = 4 S’s ◦ 2! = 2 P’s

 11*10*9*8*7*6*5*4! / 4!*4*3*2*1*2*1 =  11 * 10 * 3 * 7 * 3 * 5 =  110 * 21 * 15 =  34,650 ways  If we didn’t do “distinguishable” ways, there are 39,916,800 permutations for arranging the letters of the name MISSISSIPPI.

 See Schoolwires for a homework page with problems related to these topics.  See me for additional copies.

 White Book: Page : #12-28 even (There are fundamental counting principle, permutation, and combination problems here. If you just divide them up, you probably won’t get practice with all the concepts.) We still have part 2 of the lesson to do…