Counting Techniques Section 5.5

Objectives Solve counting problems using the Multiplication Rule Solve counting problems using permutations Solve counting problems using combinations Solve counting problems involving permutations with non-distinct items Compute probabilities involving permutations and combinations

Solve Counting Problems Using the Multiplication Rule The fixed price dinner at Applebees provides the following choices: – Appetizer: soup or salad – Entrée: baked chicken, broiled beef patty, baby beef liver or roast bee au jus – Dessert: ice cream or cheesecake How many different meals can be ordered?

How Many Meals? Appetizer – 2 choices Entree – 4 choices Dessert – 2 choices Best way to figure out the different meals is to draw them out.

How Many Meals? DessertEntreeAppetizer SoupBeefIce CreamCheesecakeLiverIce CreamCheesecakeChickenIce CreamCheesecakePattyIce CreamCheesecake DessertEntreeAppetizer SaladBeefIce CreamCheesecakeLiverIce CreamCheesecakeChickenIce CreamCheesecakePattyIce CreamCheesecake There are 8 different meals that can be constructed for a meal that start with a soup and 8 more for those that start with a salad. 16 different meals total Notice: 2 * 4 * 2 = 16

Multiplication Rule of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second, r selections for the third choices, and so on...then the task of making these selections can be done in p * q * r * …. different ways.

Example If a state has a license plate with 3 letters and 3 single digits, how many different license plates can be make? # plates = 26 * 26 * 26 * 10 * 10 * 10 = 17,576,000

Counting Without Repetition Three members from a 14 member committee are to be randomly selected to serve as chair, vice-chair and secretary. The first person selected is the chair; the second the vice-chair; the third the secretary. How many different committees can be formed? = 14 * 13 * 12 = 2184 (Hint: after the 1 st person is selected, only 13 are left to select from. After the 2 nd person is selected only 12 are left to select from)

Factorial Symbol/Function When we multiply by continuously decreasing integers greater than or equal to zero, we have a function (yes, on our calculators) to make this easier and take less space. – Called the Factorial function…. 0! = 1 1! = 1 2! = 2 * 1 3! = 3 * 2 * 1 n! = n (n – 1) * … * 3 * 2 * 1

Factorials You can do math with factorials as well, just remember what the “!” stands for. – Ex: 4! / 2! = 4 * 3 = 12

Solving Problems Using Permutations Permutation – an ORDERED arrangement in with r objects are chosen from n distinct (different) objects and repetition is not allowed. – The symbol n P r represents the number of permutations of r objects selected from n objects. – Example: Three members from a 14 member committee are to be randomly selected to serve as chair, vice-chair and secretary. The first person selected is the chair; the second the vice-chair; the third the secretary. How many different committees can be formed? Can be solved with permutations. 14 items, taken 3 at a time and order matters. 14 P 3 = 14 * 13 * 12 = 2184

Permutations - definition Number of Permutations of r objects chosen from n objects, in which: – The n objects are distinct – Once an object is used it cannot be repeated – Order is IMPORTANT Is given by the formula:

Permutation Examples Evaluate: 7 P 5 = 7! = 7*6*5*4*3*2! = 7*6*5*4*3 = 2520 (7-5)! 2! 8 P 2 = 8! = 8*7*6! = 8*7 = 56 (8-2)! 6! 5 P 5 = 5! = 5! = 5*4*3*2*1 = 120 (5-5)! 0!

Example In how many ways can 20 cars in a race finish first, second and third? 20 P 3 = 20! = 20! = 20*19*18*17! = 20*19*18 = 6840 (20-3)! 17! 17!

Solving Problems Using Combinations For permutations, ORDER IS IMPORTANT – For instance: ABC, CAB, BCA, are all different arrangements For combinations, order is NOT important. – For instance: ABC, CAB, BCA are all the SAME arrangement.

Combinations - Definition A combination is a collection, without regard to order, of n distinct objects without repetition. The symbol n C r represents the number of combinations of n distinct objects taken r at a time.

Combinations - Definition Number of different arrangements of of n objects using r<n of them, in which: – The n objects are distinct – Once an object is used it cannot be repeated – Order is NOT IMPORTANT Is given by the formula:

Example Jace, Holly, Alessandra and Marco are all going to play golf. They will randomly select teams of two player each. How many team combinations are possible? (Note: a team of Jace/Holly is the same as Holly/Jace…order is not important). 4 C 2 = 4! = 4! = 4*3*2! = 12 = 6 2!(4-2)! 2!*2! 2*1 * 2! 2

Combination Examples Evaluate: 4 C 1 = 4! = 4! = 4*3!= 4 1!(4-1)! 1! * 3! 1*3! 6 C 4 = 6! = 6! = 6*5*4! = 6 * 5 = 30 = 15 4!(6-4)! 4!*2! 4! * 2! 2*1 2 6 C 2 = 6! = 6! = 6*5*4! = 6 * 5 = 30 = 15 2!(6-2)! 2!*4! 2! * 4! 2*1 2

Calculator Functions Factorial: in MATH->PRB menu Enter number first, select function Permutations/Combination s: in MATH->PRB menu Enter “n”, select function, enter “r”

Solve Counting Problems Involving Permutations with Non-Distinct Items Sometimes we need to arrange objects in order, but some of the objects are not distinguishable.

Example Example: How many distinguishable strings of letters can be formed by using all the letters in the word REARRANGE? – Choose the positions for the 3 R’s Order matters so, can be done in 9 C 3 ways Then there are 6 positions left to be filled – Choose the positions for the 2 A’s Order matters so, can be done in 6 C 2 ways Then there are 4 positions left to be filled – Choose the positions for the 2 E’s Order matters so, can be done in 4 C 2 ways Then there are 2 positions left to be filled – Choose the position for the 1 N Order matters so, can be done in 2 C 1 ways Then there are 1 position left to be filled – Choose the position for the 1 G Order matters so, can be done in 1 C 1 ways

Example Cont’d By the Multiplication Rule, the number of possible words that can be formed is: 9 C 3 * 6 C 2 * 4 C 2 * 2 C 1 * 1 C 1 = = 15, 120 Had each of the letters in “REARRANGE” been different, there would have been 9 C 9 possible words formed.

Permutations with Non-Distinct Items The number of permutations of n objects of which n 1 are one of a kind, n 2 are of a second kind, …., and n k are of a kth kind is given by: n!_____ n 1 !*n 2 !*…*n k ! Where n = n 1 +n 2 +…+n k

Figuring Out Which Method to Use Are you figuring out the probability of a compound event? Are you figuring out a sequence of choices?

Summary

Which One Do I Use?

Assignment Pg : 1, 2, 3, 5, 7, 9, 12, 13, 15, 18, 20, 22, 23, 24, 31, 34, 51, 52 Pg 305 – 306: 60, 64, 67