Section 7.2, Part 1Combinations, 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)

Fundamental Counting Principle

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”.

Answer

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)

Factorials !!!!

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!

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!

Combinations

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.

Going back to the ice cream…

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.

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.

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

Distinguishable Permutations

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.

The answer…

See Schoolwires for a homework page with problems related to these topics.

See me for additional copies.

Homework

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…

Group Work