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Section 7.2, Part 1

Section 7.2, Part 1. Combinations, Permutations, and the Fundamental Counting Principle. 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.

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Section 7.2, Part 1

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  1. Section 7.2, Part 1 Combinations, Permutations, and the Fundamental Counting Principle

  2. 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)

  3. Answer • 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”.

  4. Factorials !!!! • 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)

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

  6. Combinations • 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? • 12C3 = 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!

  7. Going back to the ice cream… • 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) • 6C2 * 3C1 * 8C3 = • 6C2 = 6! / (6-2)! 2! = 6*5*4! / 4!*2*1 = 15 • 3C1 = 3! / (3-1)! 1! = 3*2! / 2!*1 = 3 • 8C3 = 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.

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

  9. Permutations • How many ways can 10 people running a race finish 1st, 2nd, and 3rd? • 10P3 = 10! / (10-3)! = 10 * 9 * 8 * 7! / 7! = 10 * 9 * 8 = 720 • There are 720 ways 10 people running a race can finish 1st, 2nd, and 3rd.

  10. Distinguishable Permutations • 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. The answer… • 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.

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

  13. Group Work • White Book: Page 157-158: #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…

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