Section 6.4 Counting and Combinations with Multiple Cases

Counting With Multiple Scenarios In this section, we will the number of outcomes when different scenarios/cases occur This happens when we have a string of events,ie: A & B, where event “A” has two or more options, and each of these options create a different number of options for event “B” Ex: Andy has three friends and he can only take one to go to the movies. If he chooses either Brad or Chris, he can watch either an “action”, “sci-fi”, or “comedy”. If he chooses Donna, he can only choose between a “comedy” or “drama”. How many ways can he go to the movies? There are two Events – Pick a friend and choose a movie We need to split this question into two cases, because depending on which friend he picks, he has different options for event 2

Permutations with Multiple Cases Here are some examples when we can have other multiple cases A four digit even number with non repeating digits The last digit will affect the number of options for the first digit Seating two people in a row of chairs that can not sit next to each other Where the first person sits (end or middle) can affect the number of options for the second person A combination lock (0 to 59) where the difference between the first and second digit is atleast 3 The first number will affect the number of options we have for the second number

Ex: There are 6 chairs in a row Ex: There are 6 chairs in a row. How many ways can we seat Adam and Betty if they can not sit next to each other? Suppose you give each seat a letter from A to F There are two cases, whether if Adam sits at the end or somewhere in the middle Case #2) Somewhere in the middle Case #1) At the end

Practice: How many four digit even numbers are possible if none of the digits can repeat? The last digit must be even to be an even number 4th Digit: The first digit can not be zero to be 4 digit number 1st Digit: So we have a problem!!! If the last digit was a zero, then we have 9 options for the first digit. If the last digit was either 2, 4, 6, or 8, then we only have 8 options for the first digit So we have 2 cases now!! Case 1: Last digit is zero Case 2: Last digit is 2, 4, 6, or 8 © Copyright all rights reserved to Homework depot: www.BCMath.ca

Ex: A lock has numbers from 0 to 59 and has a 3 number combination Ex: A lock has numbers from 0 to 59 and has a 3 number combination. The difference between the 1st and 2nd number is atleast 3. How many different combinations can there be?

Counting What you don’t Want Sometimes, the best way to count what you want is by counting what you don’t want, and then subtracting it from all the possible outcomes Ex: How many 3 digit numbers have atleast 2 digits that are the same? How many 3 digit numbers are there? How many 3 digit numbers have digits that are all different? Subtract them to find how many numbers that have at least 2 digit similar digits

Ex: How many triangles can you make with these 25 dots? How many ways can you select 3 points from 25? How many ways can you select 3 points that does not make a triangle? If you select 3 points from a line, it doesn’t make a triangle There are 5 horizontal lines and 5 vertical lines Also subtract 3 points that make a diagonal line So the number of triangles will be:

Challenge: How many Social Insurance Numbers (SIN) can be issued by the Government of Canada if: a) the use of “0” is not allowed in the first 3 digits, (b) no digit is allowed to be used more than 5 times, and (c) each SIN consist of 8 digits.

The Bucket Method: When we are choosing from two or more different groups with NO overlapping outcomes, we can use the “Bucket” method Ex: A group of 10 students has 4 boys and 6 girls. If Four people are chosen to form a team, how many different teams can be created if there must be 2 boys. In this scenario, there are two different groups: Boys vs Girls In a team of 4, if there are 2 boys, then there must also be 2 girls So draw a “Bucket” to separate the two groups

Some situations involve several cases: Practice: If there are 7 boys & 10 girls and 6 people are chosen to form a team, how many different teams can be created if there must be 2 Girls. Some situations involve several cases: Ie: Teams with atleast 2 girls or atmost 2 girls “atleast” means you can have 2 or more “atmost” means you can have 2 or less We need to add all the different cases with different number of girls © Copyright all rights reserved to Homework depot: www.BCMath.ca

Ex: If there are 4 boys & 6 girls, and 4 people are chosen to make a team, how many different combinations can be created if there must be ATLEAST 1 boy in the team? Must consider all the different cases with: 1 boy & 3 girls, 2 boys & 2 girls, 3 boys & 1 girl , 4 boys & 0 girls Practice: A team is to be formed with 6 teachers. If there are 5 math, 8 Science, & 3 English teachers, how many different teams can be formed if there must be atleast 3 math teachers? Must consider all the different cases with: 3 Math & 3 Others + 4 Math & 2 Others + 5 Math & 1 Other © Copyright all rights reserved to Homework depot: www.BCMath.ca

Practice: a) How many 5 card hands contain no aces and atleast 3 royal cards? Aces are not included, then we have 48 cards left Need to consider the hands with 3, 4, and 5 royal cards b) How many hands with atleast 3 spades? Need to consider the hands with 3, 4, and 5 spades © Copyright all rights reserved to Homework depot: www.BCMath.ca

Challenge: When given a five card hand, how many full house hands can you get? ( 3 of a kind and a pair) Note: there are 13 different values to choose for the pair and 3 of a kind Out of the 13 values, there are 13 choices for the triple Then there are 12 values left for the pair For each value, there are 4 suits: (Spade, hearts, diamonds and clubs) For the three of a kinds, out of the 4 suits, choose 3 For the pair, out of the 4 suits, choose 3 © Copyright all rights reserved to Homework depot: www.BCMath.ca

Out of the 13 choices, choose 2 for the pair Challenge: When dealt a five card hand, what is the probability of getting 2 pairs? Out of the 13 choices, choose 2 for the pair Draw a bucket to separate each group Out of the deck of 52 cards, 44 cards are left for the single card because it can not be the same as either one of the pairs © Copyright all rights reserved to Homework depot: www.BCMath.ca