Section 1:
DAY ONE: Examples 1  6 (34 minutes) 

Lecture 1 

06:27 

Example 1: Introduction to Combinations  There are four marbles on a table, and each marble is a different color (red, green, blue, and yellow). Two marbles are selected from the table at random and put in a bag. Is the order of the marbles, or the order of their colors, important?
 Let\'s take a moment to discuss the difference between permutations and combinations.
 The term permutation means arrangement, and order matters.
 The term combination means selection, and order does not matter.
 In word problems, it is up to you to determine if the context of a problem requires an arrangement (permutation) or a selection (combination).
 In this question, we have four marbles with different colors, and we put two in a bag.
 The order of the colors is not important. If we open the bag and look at the marbles, there is no reason why the order of the marbles or the order of their colors should matter.
 In Part B, use a tree diagram to find the number of unique color combinations.
 If the first marble is red, the second marble can be green, blue, or yellow.
 If the first marble is green, the second marble can be red, blue, or yellow.
 If the first marble is blue, the second marble can be green, red, or yellow.
 If the first marble is yellow, the second marble can be green, blue, or red.
 The result tells us that there are twelve unique arrangements. However, we want the number of unique selections, not arrangements!
 Remove arrangements that have duplicate colors.
 There is already a red and green selection, so remove the arrangement with duplicate colors.
 There is already a red and blue selection, so remove the arrangement with duplicate colors.
 There is already a red and yellow selection, so remove the arrangement with duplicate colors.
 There is already a green and blue selection, so remove the arrangement with duplicate colors.
 There is already a green and yellow selection, so remove the arrangement with duplicate colors.
 There is already a blue and yellow selection, so remove the arrangement with duplicate colors.
 There are six unique color combinations.
 In Part C, use combination notation to find the number of unique color combinations.
 A combination is represented with the notation nCr, where n is the number of items in our set, and r is the number of items to be selected.
 In this example, we can select any two marbles out of the four available marbles using the notation 4C2.
 4C2 can be evaluated in a TI83 calculator with the input sequence 4  MATH  Left Arrow  nCr  2. The result is 6, the same as in part b.
 In Part D, what is meant by the terms singlecase combination and multicase combination?
 Singlecase and multicase combinations are treated similar to singlecase and multicase permutations.
 A singlecase combination occurs when we select items from a set with only one draw from the set.
 A multicase combination occurs when we select items from a set with more than one draw from the set.
 In Part E, how many ways can three or four marbles be chosen?
 Set up the sample set with the four marbles.
 In our first case, we will select three marbles. This can be done in 4C3 = 4 ways.
 Restock and Redraw.
 In our second case, we will select four marbles. This can be done in 4C4 = 1 way.
 Add the combinations from each case to get the answer of 5.


Lecture 2 

04:52 

Example 2: Sample Sets with No Subdivisions  In this example we will study sample sets with no subdivisions. In Part A, there are five toppings available for a pizza (mushrooms, onions, pineapple, spinach, and tomatoes). If a pizza is ordered with three toppings, and no topping may be repeated, how many different pizzas can be created?
 There are five toppings available to choose from. In our sample set, use tiles to represent each topping.
 This question is a combination because order does not matter when spreading toppings on a pizza.
 We can select three toppings for the pizza in 5C3 = 10 ways.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 We have a small sample set, so it is possible to count out the 10 possible combinations.
 In Part B, a committee of 4 people is to be formed from a selection pool of 9 people. How many possible committees can be formed?
 There are nine people available to choose from. In our sample set, use tiles to represent each person.
 This question is a combination because the committee has no titled positions. All we want to do is create a group of four people from a selection pool of nine people. If there were titled positions in the committee, like president and treasurer, then order would matter and this would be a permutation.
 We can select four people for the committee in 9C4 = 126 ways.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part C, how many 5card hands can be made from a standard deck of 52 cards?
 The 52 cards in a standard deck are shown.
 This question is a combination. When we speak of a 5card hand, we are only interested in which specific cards are in the hand. The way a player orders the cards in their hand is irrelevant.
 We can select five cards in 52C5 = 2598960 ways.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part D, there are 9 dots randomly placed on a circle. i) How many lines can be formed within the circle by connecting two dots? ii) How many triangles can be formed within the circle?
 We can form a line by connecting two dots. If we call two particular dots A and B, we get the same line if we draw AB or BA.
 The order of the dots selected for the line does not matter, so this is a combination.
 We can form 9C2 = 36 different lines.
 Now we\'ll move on to the second question.
 We can form a triangle by connecting any three dots. If we call three particular dots A, B, and C, we get the same triangle no matter how we connect the lines. Therefore, triangle construction is a combination.
 We can form 9C3 = 84 different triangles.


Lecture 3 

05:46 

Example 3: Sample Sets with Subdivisions  In this example, we will study sample sets with subdivisions. In Part A, how many 6person committees can be formed from 11 men and 9 women, if 3 men and 3 women must be on the committee?
 There are 20 people in our sample set, subdivided into 11 men and 9 women.
 Now we\'ll select people for the committee.
 We can select three men for the committee in 11C3 = 165 ways.
 We can select three women for the committee in 9C3 = 84 ways.
 There are 11C3 × 9C3 = 13860 ways to form the sixperson committee.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part B, a crate of toy cars contains 10 working cars and 4 defective cars. How many ways can 5 cars be selected if only 3 work?
 There are 14 toy cars in our sample set, subdivided into 10 working cars and 4 defective cars.
 Now we\'ll select the cars.
 We can select three working toy cars in 10C3 = 120 ways.
 We can select two defective toy cars in 4C2 = 6 ways.
 There are 10C3 × 4C2 = 720 ways to select the five cars.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 Now we\'ll move on to Part C. From a deck of 52 cards, a 6card hand is dealt. How many distinct hands are there if the hand must contain 2 spades and 3 diamonds?
 Subdivide the 52 cards into: 13 spades, 13 diamonds, and 26 other cards.
 Now we\'ll select cards to form a sixcard hand.
 We can select two spades in 13C2 = 78 ways.
 We can select three diamonds in 13C3 = 286 ways.
 We can select one other card in 26C1 = 26 ways.
 The number of sixcard hands is 13C2 × 13C3 × 26C1 = 580008.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part D, a bouquet contains four types of flowers. A florist is making a bouquet that uses one type of focal flower, no fragrant flowers, three types of line flowers and all of the filler flowers. How many different bouquets can be made?
 Bring up the flower chart.
 Use tiles to represent each type of flower.
 Now we\'ll select flowers for the bouquet.
 We can select one focal flower in 6C1 = 6 ways.
 We can select no fragrant flowers in 7C0 = 1 way.
 We can select three line flowers in 5C3 = 10 ways.
 We can select all the filler flowers in 5C5 = 1 way.
 There are 6C1 × 7C0 × 5C3 × 5C5 = 60 ways to form a bouquet.


Lecture 4 

08:13 

Example 4: More Sample Sets with Subdivisions  In this example, we will study more sample sets with subdivisions. In Part A, a committee of 5 people is to be formed from a selection pool of 12 people. If Carmen must be on the committee, how many unique committees can be formed?
 Begin with a sample set of 12 people.
 Restructure the sample set to separate the people who must be on the committee from the others.
 Only one person, Carmen, must be on the committee.
 The rest of the people are in the general selection pool.
 Now we\'ll make our selections for the committee.
 Carmen must be on the committee, and she can be selected in 1C1 = 1 way.
 Four more people can be selected for the committee in 11C4 = 330 ways.
 There are 1C1 × 11C4 = 330 ways to form the fiveperson committee.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part B, a committee of 6 people is to be formed from a selection pool of 11 people. If Grant and Helen must be on the committee, but Aaron must not be on the committee, how many unique committees can be formed?
 Begin with a sample set of 11 people.
 Restructure the sample set to separate the constraints.
 Grant and Helen must be on the committee.
 Aaron must NOT be on the committee.
 And everyone else remains in the general selection pool.
 Now we\'ll make our selections for the committee.
 Grant and Helen must be on the committee, and they can be selected in 2C2 = 1 way.
 Four additional people can be selected in 8C4 = 70 ways.
 There are 2C2 × 8C4 = 70 ways to form the sixperson committee.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part C, nine students are split into three equalsized groups to work on a collaborative assignment. How many ways can this be done?
 Begin with a sample set of 9 people.
 We are going to create three groups, each with three people.
 The first group can be selected in 9C3 = 84 ways.
 The second group can be selected in 6C3 = 20 ways.
 The third group can be selected in 3C3 = 1 way.
 There are 9C3 × 6C3 × 3C3 = 1680 ways to form the three groups.
 No constraints were placed on the selections, so there was no need to subdivide the sample set.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 Now we\'ll move on to Part D. From a deck of 52 cards, a 5card hand is dealt. How many distinct 5card hands are there if the ace of spades and two of diamonds must be in the hand?
 The 52 cards in a standard deck are shown.
 Separate the cards that must be included from the rest of the cards.
 The two cards that must be included can be selected in 2C2 = 1 way.
 Now select three other cards in 50C3 = 19600 ways.
 We can select the five cards in 2C2 × 50C3 = 19600 ways.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.
 In Part E, a lottery ticket has 6 numbers from 149. Duplicate numbers are not allowed, and the order of the numbers does not matter. How many different lottery tickets contain the numbers 12, 24 and 48, but exclude the numbers 30 and 40?
 List the available numbers.
 Restructure the sample set to separate the constraints.
 12, 24, and 48 must be on the ticket.
 30 and 40 must NOT be on the ticket.
 The rest of the numbers can go in the general selection pool.
 Now we\'ll make our selections for the lottery ticket.
 12, 24, and 48 must be on the lottery ticket, and they can be selected in 3C3 = 1 way.
 Select three other numbers in 44C3 = 13244 ways.
 There are 3C3 × 44C3 = 13244 ways to select numbers for the lottery ticket with the given constraints.
 Note that this is a singlecase combination because we do not restock and redraw from the sample set.


Lecture 5 

04:51 

Example 5: Permutations and Combinations Together  In this example, we will study questions involving both permutations and combinations. In Part A, how many fiveletter words using letters from TRIANGLE can be made if the fiveletter word must have two vowels and three consonants?
 Let\'s outline a strategy to solve this question. First, find the number of ways we can combine two vowels and three consonants from TRIANGLE.
 And second, each unordered combination of letters needs to be ordered into words.
 Subdivide the letters of TRIANGLE into consonants and vowels.
 Replace the letters with tiles before we continue.
 Now we\'ll select letters to form a fiveletter word.
 We can select three consonants for the word in 5C3 = 10 ways.
 We can select two vowels for the word in 3C2 = 3 ways.
 We can arrange the five letters we selected in 5P5 = 5! ways.
 The answer is 5C3 × 3C2 × 5! = 3600.
 Now we\'ll move on to Part B. There are 4 men and 5 women on a committee selection pool. A threeperson committee consisting of President, VicePresident, and Treasurer is to be formed. How many ways can exactly two men be on the committee?
 We have nine people, subdivided into 4 men and 5 women.
 Now we\'ll select people to form a committee.
 We can select two men for the committee in 4C2 = 6 ways.
 We can select one woman for the committee in 5C1 = 5 ways.
 The committee has titles for each position, so order matters in a committee.
 We can order the three committee members in 3P3 = 3! ways.
 The answer is 4C2 × 5C1 × 3! = 180.
 Now we\'ll move on to Part C. A music teacher is organizing a concert for her students. If there are six piano students and seven violin students, how many different concert programs are possible if four piano students and three violin students perform in an alternating arrangement?
 We have 13 students subdivided into 6 piano students and 7 violin students.
 Now select piano students and violin students for the concert.
 We can select four piano students for the concert in 6C4 = 15 ways.
 We can select three violin students for the concert in 7C3 = 35 ways.
 Now arrange the program so violin and piano performances alternate.
 Using the fundamental counting principle, there are 144 arrangements.
 The answer is 6C4 × 7C3 × 144 = 75600.


Lecture 6 

04:01 

Example 6: Handshakes, Teams, and Shapes  In this example, we will study handshakes, teams, and shapes. In Part A, twelve people at a party shake hands once with everyone else in the room. How many handshakes took place?
 The handshake problem will have an identical solution to "How many lines can be formed from 12 dots on a circle?"
 In this context, each dot represents a person, and the line connecting them represents a handshake.
 We can form 12C2 = 66 lines, so there are 66 handshakes.
 Now we\'ll move on to Part B. If each of the 8 teams in a league must play each other three times, how many games will be played?
 We can begin this problem by answering the similar problem: "How many lines can be formed from 8 dots on a circle?"
 In this context, each dot represents a team and the line connecting them represents a game.
 Each team plays each other once 8C2 = 28 times.
 Each team plays each other a second time 8C2 = 28 times.
 And each team plays each other a third time 8C2 = 28 times.
 Add the total number of games to get the answer, 84.
 Note: We are adding the results of three identical cases, so technically this question should be classified as a multicase combination.
 Now we\'ll move on to Part C. If there are 8 dots on a circle, how many quadrilaterals can be formed?
 If we draw eight dots on a circle, we can form a quadrilateral by choosing any four of the dots.
 We can form 8C4 = 70 quadrilaterals.
 Now we\'ll move on to Part D. A polygon has 6 sides. How many diagonals can be formed?
 A diagonal is formed by joining two nonconsecutive vertices of a polygon.
 By this definition, the sides of the polygon are not diagonals.
 We can join any two vertices in 6C2 = 15 ways.
 However, six of the lines formed will be the sides of the polygon, so they must be subtracted from the total. There are 6C2  6 = 9 diagonals that can be formed.
 In general, the number of diagonals in a polygon is nC2  n, where n is the number of sides or vertices in the polygon.

Section 2:
DAY TWO: Examples 7  11 (32 minutes) 

Lecture 7 

04:32 

Example 7: Combinations with Repetitions Allowed  In this example, we will study combinations where repetitions are allowed. In Part A, a jar contains quarters, loonies, and toonies. If four coins are selected from the jar, how many unique coin combinations are there?
 In this question, we know there are quarters, loonies, and toonies in the jar, but no quantity has been specified!
 Since no quantity has been specified, and it is possible to select the same type of coin more than once, we can treat the coins as being unlimited.
 Some of the ways we can choose four coins include: LQTL
 Another possible combination is QQQL.
 And another possible combination is TTTT.
 We need a new type of strategy to deal with this problem.
 Create a chart to keep track of the number of quarters, loonies, and toonies selected.
 Draw checkmarks and dividers for LQTL.
 Draw checkmarks and dividers for QQQL.
 And draw checkmarks and dividers for TTTT.
 Notice that we can represent each combination as an arrangement of four identical checkmarks and two identical dividers.
 We can arrange the four identical checkmarks and two identical dividers in: 6!/(4!2!) = 15 ways.
 If we actually count out the number of unique combinations, we get 15.
 In Part B, a bag contains marbles with four different colors (red, green, blue, and yellow).
If three marbles are selected from the bag, how many unique color combinations are there?  In this question, we know there marbles with different colors in a bag, but no quantity has been specified!
 Since no quantity has been specified, and it is possible to select the same type of marble more than once, we can treat the marbles as being unlimited.
 Create a chart to keep track of the number of marbles selected.
 One possible selection is RYY.
 Another possible selection is GGG.
 And another possible selection is GBY.
 Notice that we can represent each combination as an arrangement of three identical checkmarks and three identical dividers.
 We can arrange the three identical checkmarks and three identical dividers in: 6!/(3!3!) = 20 ways.
 If we actually count out the number of unique combinations, we get 20.


Lecture 8 

10:53 

Example 8: MultiCase Combinations  In this example, we will study combinations with more than one case. In Part A, a committee of 5 people is to be formed from a group of 4 men and 5 women. How many committees can be formed if at least 3 women are on the committee?
 There are 9 people available for selection, subdivided into 4 men and 5 women.
 There are three ways we can form the committee: 3 women & 2 men, 4 women & 1 man, and 5 women & 0 men.
 We can select 3 women and 2 men in 5C3 × 4C2 = 60 ways.
 Restock and Redraw.
 We can select 4 women and 1 man in 5C4 × 4C1 = 20 ways.
 Restock and Redraw.
 We can select 5 women in 5C5 = 1 way.
 Add the cases to get the answer, 81.
 This is a multicase combination because we restock and redraw the sample set and add the results.
 Now we\'ll move on to Part B. From a deck of 52 cards, a 5card hand is dealt. How many distinct hands can be formed if there are at most 2 queens?
 Begin with the standard deck of 52 cards.
 Separate the queens from the deck.
 There are three cases: 0 queens and 5 other cards, 1 queen and 4 other cards, and 2 queens and 3 other cards.
 We can select 0 queens and 5 other cards in 48C5 ways.
 Restock and Redraw.
 We can select 1 queen and 4 other cards in 4C1 × 48C4 ways.
 Restock and Redraw.
 We can select 2 queens and 3 other cards in 4C2 × 48C3 ways.
 Add the cases to get 2594400.
 This is a multicase combination because we restock and redraw the sample set and add the results.
 Now we\'ll move on to Part C. From a deck of 52 cards, a 5card hand is dealt. How many distinct hands can be formed if there is at least 1 red card?
 Begin with the standard deck of 52 cards.
 If there is at least one red card, we have five cases: 1 red, 4 black; 2 red, 3 black; 3 red, 2 black; 4 red, 1 black; and 5 red, 0 black.
 While we could do the question this way, using five cases, there is a shortcut!
 In total, there are six cases possible for selecting the cards:
 If at least one red card must be in the hand, we don\'t want the case with zero red cards.
 So, we can find the answer using: Total Combinations – Unwanted Combinations.
 The total cases are any five cards, and the unwanted case is 0 red cards and 5 black cards.
 We can select any five cards in 52C5 ways.
 Restock the cards.
 Now subdivide the deck into black cards and red cards.
 We can select all black cards in 26C5 ways.
 Subtract. There are 2533180 ways to select at least one red card.
 Now we\'ll move on to Part D. A research team of 5 people is to be formed from 3 biologists, 5 chemists, 4 engineers, and 2 programmers. How many teams have exactly one chemist and at least 2 engineers?
 Represent the people in the selection pool with tiles. The biologists and programmers can be grouped as "others" because our constraints do not specifically use either profession.
 There are three cases: 1 chemist, 2 engineers, and 2 others; 1 chemist, 3 engineers, and 1 other; and 1 chemist and 4 engineers.
 We can select 1 chemist, 2 engineers, and 2 others in: in 5C1 × 4C2 × 5C2 ways.
 Restock and Redraw.
 We can select 1 chemist, 3 engineers, and 1 other in: in 5C1 × 4C3 × 5C1 ways.
 Restock and Redraw.
 We can select 1 chemist and 4 engineers in 5C1 × 4C4 ways.
 Add the results to get 405 ways to form the research team.
 Now we\'ll move on to Part E. In how many ways can you choose one or more of 5 different candies?
 The sample set contains five candies.
 There are five cases: 1 candy selected, 2 candies selected; 3 candies selected, 4 candies selected; or 5 candies selected.
 We can choose 1 candy in 5C1 ways.
 Restock and Redraw.
 We can choose 2 candies in 5C2 ways.
 Restock and Redraw.
 We can choose 3 candies in 5C3 ways.
 Restock and Redraw.
 We can choose 4 candies in 5C4 ways.
 Restock and Redraw.
 We can choose 5 candies in 5C5 ways.
 Add the five cases to get 31.
 While the method we just used works, it would become tedious for large sample sets. Let\'s look at another method that can be used to solve this problem.
 Each candy has two states associated with it: it can either be selected or not selected.
 There are two options for the first candy (yes or no). We can select one option in 2C1 ways.
 All of the other candies can either be selected or not selected in 2C1 ways as well.
 Multiply the results to get (2C1)^5 = 25 = 32.
 We don\'t want to include the case where no candy is selected, so subtract 1 from the total. (recall that the question specifies one or more).
 The answer is 32  1 = 31.
 We can generalize these types of oneormore questions with the formula: Combinations = 2^n – 1, where n is the number of objects that can be selected.


Lecture 9 

05:01 

Example 9: Combination Formula  In this example we will study the combination formula. In Part A, evaluate 7C5.
 The formula for a combination is: nCr = n!/(nr)!r!, where n is the number of items in the selection pool, and r is the number of items to be chosen.
 Rewrite the combination formula.
 Plug in n and r.
 Simplify.
 Expand 7!. Stop expanding at 5! since this will cancel out the 5! in the denominator.
 Cancel the 5!\'s.
 The answer is 21.
 In Part B, evaluate 3C3.
 Write the combination formula.
 Plug in n and r.
 Simplify.
 0! = 1.
 The 3!\'s cancel, giving the answer 1.
 In Part C, evaluate 4C2.
 n over r in brackets is sometimes used instead of nCr to represent a combination. The formula remains the same: nCr = n!/(nr)!r!.
 Rewrite the combination formula.
 Plug in n and r.
 Simplify.
 Expand 4!. Stop expanding at 2! since this will cancel out the 2! in the denominator.
 Cancel the 2!\'s.
 The answer is 6.
 In Part D, write 6!/(4!2!) as a combination.
 Write the combination formula.
 Rewrite the expression on the right side.
 By comparison with the combination formula, we can identify n = 6 and r = 2.
 The answer is 6C2.
 In Part E, write 5!/4! as a combination.
 Write the combination formula.
 Rewrite the expression on the right side.
 By comparison with the combination formula, we can identify n = 5, but we can\'t identify r.
 By inspection, r = 1.
 We need to get 4! in the denominator. Using guessandcheck, we get 4! when r = 1.
 The answer is 5C1.


Lecture 10 

06:23 

Example 10: Equations with Combinations  In this example we will continue to study the combination formula. Solve for the unknown algebraically. In Part A, find n in nC2 = 21.
 Expand the left side using the combination formula.
 Expand the numerator. Stop at (n – 2)! since this cancels out the (n – 2)! in the denominator.
 Cancel the (n – 2)!\'s.
 Crossmultiply.
 Expand and bring all terms to the left side.
 Factor the trinomial.
 The solutions to the equation are n = 6 and 7.
 Reject 6 because we can\'t use negative numbers in combinations.
 The answer is n = 7.
 In Part B, find r in 4Cr = 6.
 Expand the left side using the combination formula.
 4! = 24.
 CrossMultiply.
 Divide both sides by 6.
 At this point, we\'re stuck. To move forwards, notice the following:
 We have a product of factorials on the right side, so can we express the left side as a product of factorials as well?
 The left side, 4, is equivalent to 2 × 2, which is the same as 2! × 2!.
 By inspection, we can see that r = 2.
 The combination is 4C2.
 In Part C, find n in nC3 = 10.
 Expand the left side using the combination formula.
 Expand the numerator. Stop at (n – 3)! Since this cancels out the (n – 3)! In the denominator.
 Cancel the (n – 3)!\'s.
 CrossMultiply.
 Expand the left side.
 Expand further and bring all the terms to the left side.
 There are two ways we can solve this: 1) Algebraically (by finding potential zeros and factoring using synthetic division), and 2) Graphing and finding the xintercepts.
 If we graph the polynomial (option 2), we get an xintercept at (5, 0).
 The answer is n = 5.
 In Part D, find n in nCn2 = 15.
 Expand the left side using the combination formula.
 Simplify.
 Simplify further.
 Expand the numerator. Stop at (n – 2)! since this cancels out the (n – 2)! in the denominator.
 Cancel the (n – 2)!\'s.
 2! = 2.
 CrossMultiply.
 Expand and bring all terms to the left side.
 Factor the trinomial.
 The solutions to the equation are n = 5 and 6.
 Reject 5 because we can\'t use negative numbers in combinations.
 The answer is n = 6.


Lecture 11 

05:28 

Example 11: Equations with Combinations  In this example we will continue to study the combination formula. Solve for the unknown algebraically. In Part A, find n in nC4/n2C2 = 1.
 Expand the numerator and denominator using the combination formula.
 Simplify.
 Multiply the numerator by the reciprocal of the denominator.
 Cancel.
 Expand the numerator. Stop at (n – 2)! since this cancels out the (n – 2)! in the denominator.
 Cancel.
 Crossmultiply.
 Bring all terms to the left side.
 Factor the trinomial.
 The solutions to the equation are n = 3 and 4.
 Reject 3 because we can\'t use negative numbers in combinations.
 The answer is n = 4.
 In Part B, find n in nCr/nCnr = 1.
 Expand the numerator and denominator using the combination formula.
 Simplify.
 Multiply the numerator by the reciprocal of the denominator.
 All terms cancel out.
 The variables cancel out completely, so the original equation holds true for all possible nvalues, where n is a natural number.
 Another way to interpret the result is that nCr = nCnr. So, as an example, all of the following are true: 8C3 = 8C5; 5C2 = 5C3; and 12C5 = 12C7.
 In Part C, find n in n1P3 = 2 × n1C2.
 Expand the permutation and combination.
 Simplify.
 Cancel.
 CrossMultiply.
 Expand the left side. Stop at (n – 4)! since it cancels with the right side.
 Cancel the (n  4)!\'s.
 The answer is n = 4.
 In Part D, find n in n+1C2 = 1/2 × n+2C3.
 Expand using the combination formula.
 Simplify.
 Simplify further.
 Crossmultiply.
 Cancel.
 Expand the right side. Stop at (n + 1)! since it cancels with the left side.
 Cancel.
 The answer is n = 4.

Section 3:
DAY THREE: Examples 12  14 (26 minutes) 

Lecture 12 

07:57 

Example 12: Permutations and Combinations  Assorted Mix I  A sixcharacter code has the pattern shown below, and the same letter or digit may be used more than once. How many unique codes can be created?
 Pause the video and organize the information about this question.
 The sample set contains 26 letters and 10 digits. There is an infinite supply of each character, and we can use the same character more than once. Lock the positions (no rearrangements) because constraints (letter or digit) have been applied to each position.
 Fill in the number of options available for each position.
 Multiply the numbers to get 6760000 possible codes.
 Now we\'ll move on to Part B. If there are 2 different parkas, 5 different scarves, and 4 different tuques, how many winter outfits can be made if an outfit consists of one type of each garment?
 Pause the video and organize the information about this question.
 The 11 items in the sample set are subdivided into 2 parkas, 5 scarves, and 4 tuques.
 We can select one garment of each type to make 40 different winter outfits.
 Now we\'ll move on to Part C. If a 5card hand is dealt from a deck of 52 cards, how many hands have at most one diamond?
 Pause the video and organize the information about this question.
 Subdivide the 52card deck into 13 diamonds and 39 other cards.
 There are two cases we want: 0 diamonds and 5 other cards; and 1 diamond and 4 other cards.
 Add the two cases to get 1645020 possible hands.
 Now we\'ll move on to Part D. If there are three cars and four motorcycles, how many ways can the vehicles park in a line such that cars and motorcycles alternate positions?
 Pause the video and organize the information about this question.
 The sample set is subdivided into three cars and four motorcycles.
 Set up positions for the vehicles such that motorcycles and cars alternate.
 Fill in the number of options available for each position.
 Multiply the numbers to get 144 alternating arrangements.
 Now we\'ll move on to Part E. Show that nCr = nCn  r.
 Rewrite the question.
 Expand both sides using the combination formula.
 Simplify.
 Rearrange the denominator so the two sides match. This shows that nCr = nCn  r .
 Now we\'ll move on to Part F. There are nine people participating in a raffle. Three $50 gift cards from the same store are to be given out as prizes. How many ways can the gift cards be awarded?
 Pause the video and organize the information about this question.
 There are nine people in the sample set.
 The gift cards are identical, so the order in which they are awarded is irrelevant.
 We can select three people for gift cards in 9C3 = 84 ways.
 Now we\'ll move on to Part G. There are nine competitors in an Olympic event. How many ways can the bronze, silver, and gold medals be awarded?
 Pause the video and organize the information about this question.
 There are nine people in the sample set.
 This question has an identical solution to: "From a group of nine people, how many ways can we arrange three in a line?"
 By the fundamental counting principle, there are 504 ways to award the medals.
 We can also answer this as a permutation. There are 9P3 = 504 ways to award the medals.
 Now we\'ll move on to Part H. A stirfry dish comes with a base of rice and the choice of five toppings: broccoli, carrots, eggplant, mushrooms, and tofu. How many different stirfry dishes can be prepared if the customer can choose zero or more toppings?
 Pause the video and organize the information about this question.
 There are five toppings in the sample set:
 In method one, we can select zero or more toppings by adding the cases with no toppings, one topping, ... five toppings.
 In method two, we can assign two states to each topping: it can either be selected, or not selected.
 We can use the formula for zeroormore combinations, 2^n, or we can multiply 2C1 five times to get the answer, 32.


Lecture 13 

09:05 

Example 13: Permutations and Combinations  Assorted Mix II  A set of tiles contains eight letters, A  H. If two of these sets are combined, how many ways can all the tiles be arranged? Leave your answer as an exact value.
 Pause the video and organize the information about this question.
 Mix the two sets of eight letter tiles together. The sample set now has 16 letters. (AABBCCDDEEFFGGHH).
 We can arrange 16 letter tiles in 16! ways. Divide out the repetitions.
 The exactvalue answer is 16!/(2!)^8.
 Now we\'ll move on to Part B. A pattern has five dots such that no three points are collinear. How many lines can be drawn if each dot is connected to every other dot?
 Pause the video and organize the information about this question.
 There are five dots, and we need to select two dots to form a line.
 There are 5C2 = 10 lines that can be formed.
 Now we\'ll move on to Part C. How many ways can the letters in CALGARY be arranged if L and G must be separated?
 Pause the video and organize the information about this question.
 First, find the total number of arrangements.
 The total number of ways we can arrange seven letters is 7!
 Next, find the number of unwanted arrangements.
 Put L and G in a container to indicate that they must be kept together.
 The number of arrangements that have L and G together is 6!×2!
 Finally, subtract the unwanted arrangements from the total arrangements.
 This gives us 7! – 6!2!.
 Don\'t forget to divide out the repetitions!
 The answer is 1800.
 Now we\'ll move on to Part D. A fiveperson committee is to be formed from 11 people. If Ron and Sara must be included, but Tracy must be excluded due to a conflict of interest, how many committees can be formed?
 Pause the video and organize the information about this question.
 The sample set can be subdivided into the groups: must be included; must be excluded; and the general selection pool.
 Ron and Sara are included in 2C2 = 1 way.
 Three other people can be included in 8C3 = 56 ways.
 Multiply to get 2C2 × 8C3 = 56 committees.
 Now we\'ll move on to Part E. Moving only South and East, how many unique pathways connect points A and C?
 Pause the video and organize the information about this question.
 A sample pathway from Point A to Point B is SESE.
 A sample pathway from Point B to Point C is EESESS.
 The sample set can be subdivided into 2 S\'s and 2 E\'s for A to B, and 3 S\'s and 3 E\'s for B to C.
 Use the fundamental counting principle to arrange the first set and divide out repetitions.
 Use the fundamental counting principle to arrange the second set and divide out repetitions.
 Multiply to get the answer, 120.
 Now we\'ll move on to Part F. How many ways can the letters in SASKATOON be arranged if the letters K and T must be kept together, and in that order?
 Pause the video and organize the information about this question.
 Put K and T in a container since they must stay together.
 The KT container counts as one item, so we can arrange eight items in 8! Ways.
 Ordinarily, we would also order K and T in 2! ways. However, since the order must remain fixed as KT, we do not require the 2!.
 We can arrange eight items in 8! ways. Divide out the repetitions to get the answer, 5040.
 Now we\'ll move on to Part G. A 5card hand is dealt from a deck of 52 cards. How many hands are possible containing at least three hearts?
 Pause the video and organize the information about this question.
 There are 52 cards, subdivided into 13 hearts and 39 other cards.
 There are three cases possible: 3 hearts and 2 other cards; 4 hearts and 1 other card; and all 5 hearts.
 Add the three cases together to get 241098 possible hands.
 Now we\'ll move on to Part H. A healthy snack contains an assortment of four vegetables. How many ways can one or more of the vegetables be selected for eating?
 Pause the video and organize the information about this question.
 One way to solve this would be to add up all the cases:
 4C1 + 4C2 + 4C3 + 4C4 = 15.
 Another way is to use the oneormore shortcut, 2^n – 1.
 Plug in 4 for the number of items to get the answer, 15.


Lecture 14 

08:46 

Example 14: Permutations and Combinations  Assorted Mix III  How many ways can the letters in EDMONTON be arranged if repetitions are not allowed?
 Pause the video and organize the information about this question.
 There are eight letters to be arranged. Watch out  there are two sets of repeated letters (2 O\'s and 2 N\'s).
 We can arrange eight letters in 8! ways. Divide out the repetitions to get the answer, 10080.
 Now we\'ll move on to Part B. A bookshelf has n fiction books and six nonfiction books. If there are 150 ways to choose two books of each type, how many fiction books are on the bookshelf?
 Pause the video and organize the information about this question.
 The sample set has an unknown number of items, subdivided into n fiction books and 6 nonfiction books.
 We are choosing four books from the sample set; two fiction books and two nonfiction books. This is a single case because we do not restock and redraw from the sample set.
 We can choose two fiction books in nC2 ways, and we can choose two nonfiction books in 6C2 ways. Multiply to get 150 ways to select two books of each type.
 6C2 = 15.
 Divide both sides by 15 to get nC2 = 10.
 Use guessandcheck in the calculator to get n = 5. There are five fiction books.
 Now we\'ll move on to Part C. How many different pathways exist between points A and D?
 Pause the video and organize the information about this question.
 The path ABD has 3 × 2 = 6 pathways, using the fundamental counting principle.
 The path ACD has 1 × 2 = 2 pathways, using the fundamental counting principle.
 Add the results of the two cases to get 8 pathways.
 Now we\'ll move on to Part D. How many numbers less than 60 can be made using only the digits 1, 5, and 8, if the numbers formed may contain repeated digits?
 Pause the video and organize the information about this question.
 Numbers less than 60 include both onedigit and twodigit numbers, so there are multiple cases to consider.
 There are three options for the onedigit number.
 The twodigit number must be less than 60, so there are two options for the first position (1 or 5). Any digit can go in the second position.
 There are six options for the twodigit number.
 Add the results: 3 + 6 = 9 numbers can be made.
 Now we\'ll move on to Part E. A particular college in Alberta has a list of approved prerequisite courses. Five courses are required for admission to the college. Math 301 (or Math 302) and English 301 are mandatory requirements, and at least one science course must be selected as well. How many different ways could a student select five courses on their college application form?
 Pause the video and organize the information about this question.
 There are three cases: 1 Math, 1 Science, 1 English, and 2 Others; 1 Math, 2 Science, 1 English, and 1 Other; and 1 Math, 3 Science, and 1 English.
 Add the cases to get 92 possible selections.
 Now we\'ll move on to Part F. How many ways can four bottles of different spices be arranged on a spice rack with holes for six spice bottles?
 Pause the video and organize the information about this question.
 We can represent the four spice bottles with the letters A, B, C, and D. The two spaces can be represented with the letters S and S. A sample arrangement is ABSSCD.
 We can arrange six items in 6! ways. There are repeating S\'s, so divide out 2!. There are 360 ways to arrange the spice bottles and spaces.
 Now we\'ll move on to Part G. If there are 8 rock songs and 9 pop songs available, how many unique playlists containing 3 rock songs and 2 pop songs are possible?
 Pause the video and organize the information about this question.
 The sample set has 17 songs, subdivided into 8 rock songs and 9 pop songs.
 We need to select five songs (3 rock and 2 pop) for the playlist. This can be done in 8C3×9C2 ways.
 Now that the five songs have been selected, arrange them into a playlist in 5! ways.
 8C3×9C2×5! = 241920 playlists.
 Now we\'ll move on to Part H. A hockey team roster contains 12 forwards, 6 defencemen, and 2 goalies. During play, only six players are allowed on the ice  3 forwards, 2 defencemen, and 1 goalie. How many different ways can the active players be selected?
 Pause the video and organize the information about this question.
 There are 20 players in the sample set, subdivided into 12 forwards, 6 defencemen, and 2 goalies.
 From the sample set, choose 3 forwards, 2 defencemen, and 1 goalie. There are 6600 ways active players can be selected.

Section 4:
DAY FOUR: Examples 15  16 (22 minutes) 

Lecture 15 

09:36 

Example 15: Permutations and Combinations  Assorted Mix IV  A fruit mix contains blueberries, grapes, mango slices, pineapple slices, and strawberries. If six pieces of fruit are selected from the fruit mix and put on a plate, how many ways can this be done?
 Pause the video and organize the information about this question.
 The fruit available for selection is not quantified, so we can assume it is infinite.
 Create a chart to form some sample fruit combinations.
 One possible combination is blueberries, one mango slice, two pineapple slices, and two strawberries.
 Another possible combination is two grapes and four pineapple slices.
 All possible combinations of six fruit can be represented as the arrangement of six identical checkmarks and four identical dividers.
 We can arrange 10 items in 10! ways. Divide out the repetitions to get 210 combinations.
 Now we\'ll move on to Part B. How many ways can six letter blocks be arranged in a pyramid, if all of the blocks are used?
 Pause the video and organize the information about this question.
 There are six blocks altogether. The sample set contains the blocks ABCDEF.
 The bottom row can be arranged in 6P3 ways, the middle row can be arranged in 3P2 ways since 3 blocks remain, and the top row can be arranged in 1P1 ways since 1 block remains.
 There are 6P3 × 3P2 × 1P1 = 720 arrangements for the pyramid.
 Note that we could have just used 6! to arrange the blocks, since drawing down a sample set using permutation notation is no different than using the fundamental counting principle.
 Now we\'ll move on to Part C. If a 5card hand is dealt from a deck of 52 cards, how many hands have cards that are all the same suit?
 Pause the video and organize the information about this question.
 There are 52 cards in the sample set, subdivided into four suits, each with 13 cards.
 There are four cases: 5 clubs; 5 diamonds; 5 spades; or 5 hearts.
 Add the four cases to get 5148 possible hands.
 Now we\'ll move on to Part D. If a 5card hand is dealt from a deck of 52 cards, how many hands have cards that are all the same color?
 Pause the video and organize the information about this question.
 There are 52 cards in the sample set, subdivided into two colors, each with 26 cards.
 There are two cases: all red cards; or all black cards.
 We can select all red cards in 26C5 ways.
 We can select all black cards in 26C5 ways.
 Add the two cases to get 131560 possible hands.
 Now we\'ll move on to Part E. A multiple choice test contains 5 questions, and each question has four possible responses. How many different answer keys are possible?
 Pause the video and organize the information about this question.
 The sample set contains four letters. We can use each letter an unlimited number of times in the answer key.
 There are four letters available for each question.
 Multiply the numbers to get 1024 possible answer keys.
 Now we\'ll move on to Part F. How many diagonals are there in a pentagon?
 Pause the video and organize the information about this question.
 Diagonals are the lines formed between two nonadjacent vertices in a polygon.
 The formula: Diagonals = nC2 – n can be used to find the number of diagonals for a polygon with n vertices.
 There are 5C2  5 = 5 diagonals that can be formed.
 Now we\'ll move on to Part G. How many ways can eight books, each covering a different subject, be arranged on a shelf such that books on biology, history, or programming are never together?
 Pause the video and organize the information about this question.
 There are eight books in the sample set, subdivided into three books that must never be together, and the other five books.
 Create positions for the five books that have no constraints. Create spaces for the books that may never go together.
 We can arrange three of the six spaces in 6P3 ways, and we can arrange the five books in 5! ways. Multiply to get 6P3 × 5! = 14400 arrangements.
 Now we\'ll move on to Part H. If a 5card hand is dealt from a deck of 52 cards, how many hands have two pairs?
 Pause the video and organize the information about this question.
 There are 52 cards in a standard deck.
 Subdivide the cards by rank.
 Now we\'ll form a sample 5card hand.
 First, select two ranks. We can do this in 13C2 ways.
 Next, select two cards from each rank.
 We may not reuse ranks that have already been chosen. There are 44 other cards remaining, and we can choose one in 44C1 way.
 The number of hands with two pairs is 13C2×4C2×4C2×44C1 = 123552.


Lecture 16 

12:18 

Example 16: Permutations and Combinations  Assorted Mix V  How many ways can six people be split into two equalsized groups?
 Pause the video and organize the information about this question.
 There are six people in the sample set.
 We can choose three people for the first group in 6C3 ways.
 We can choose three people for the second group in 3C3 ways.
 There are 6C3 × 3C3 = 20 ways to form two groups from six people.
 Now we\'ll move on to Part B. Show that 25! + 26! = 27 × 25!
 Expand 26!. Stop at 25! since this matches the first term.
 Factor out 25!.
 1 + 26 = 27.
 We have now shown that 25! + 26! = 27 × 25!.
 Now we\'ll move on to Part C. Five different types of fruit and six different types of vegetables are available for a healthy snack tray. The snack tray is to contain two fruits and three vegetables. How many different snack trays can be made if blueberries or carrots must be served, but not both together?
 Pause the video and organize the information about this question.
 There are 11 food items, subdivided into 5 fruits and 6 vegetables.
 First, create a snack tray including blueberries and excluding carrots.
 We can select blueberries in 1C1 way, another fruit in 4C1 ways, and three vegetables (excluding carrots) in 5C3 ways.
 Restock and redraw.
 Next, create a snack tray excluding blueberries and including carrots.
 We can select two fruits in 4C2 ways, carrots in 1C1 way, and two other vegetables in 5C2 ways.
 Add the two cases to get 100 possible snack trays.
 Now we\'ll move on to Part D. In genetics, a codon is a sequence of three letters that specifies a particular amino acid. A fragment of a particular protein yields the amino acid sequence: Met  Gly  Ser  Arg  Cys  Gly. How many unique codon arrangements could yield this amino acid sequence?
 Rewrite the amino acid sequence.
 There is one codon available for the first amino acid, and we require 1. This can be done in 1C1 = 1 way.
 Write the combinations for the other amino acids in the sequence.
 Condense the expression.
 Evaluate to get 1152 possible codon arrangements.
 In method two, we can use the fundamental counting principle.
 Draw six positions. We can\'t rearrange the amino acid sequence, so the positions are locked and repetitions are allowed.
 Multiply the number of options available for each position to get the answer, 1152.
 Now we\'ll move on to Part E. In a tournament, each player plays every other player twice. If there are 56 games, how many people are in the tournament?
 Pause the video and organize the information about this question.
 We can represent the problem with the equation 2 × nC2 = 56.
 Divide both sides by 2.
 Expand the left side using the combination formula.
 2! = 2.
 Crossmultiply.
 Expand the left side. Stop at (n  2)! Since this matches the right side.
 Cancel.
 Expand and bring all terms to the left side.
 Factor.
 The solution is n = 7 or 8.
 Reject 7. The answer is n = 8.
 Now we\'ll move on to Part F. The discount shelf in a bookstore has a variety of books on computers, history, music, and travel. The bookstore is running a promotion where any five books from the discount shelf can be purchased for $20. How many ways can the five books be purchased?
 Pause the video and organize the information about this question.
 The quantity of books on the discount shelf was not given, so we can assume there is an unlimited supply.
 Create a chart to form some sample book combinations.
 One possible selection is CHMTT.
 Another possible selection is HHHMM.
 We can arrange five identical checkmarks and three identical dividers in 56 ways.
 Now we\'ll move on to Part G. Show that nCr + nCr + 1 = n + 1Cr + 1.
 Rewrite the equation.
 Expand each combination on the left side using the combination formula.
 Simplify the denominator in the second term.
 Factor out n!.
 Now get (n  r  1)! in both denominators by expanding (n  r)! in the left term.
 Factor out (n  r  1)!.
 Now get r! in both denominators by expanding (r + 1)! in the right term.
 Factor out r!.
 Get a common denominator in the brackets.
 Add the fractions.
 Simplify.
 Multiply the fractions. Arrange the factors so larger factors are on the left.
 The factors in each rectangle can be condensed.
 By the combination formula, we can write the fraction as n+1Cr+1.
 Now we\'ll move on to Part H. How many pathways are there from point A to point C, passing through point B? Each step of the pathway must be getting closer to point C.
 In this 3D pathway, we can move from Point A to Point C using arrangements of North, East, and Up.
 Example pathways for A to B include: ENU.
 and UEN.
 So, the number of pathways from A to B is the same as the number of ways E, N, and U can be arranged.
 We can arrange three different letters in 3! = 6 ways.
 Since there are two cubes (we are moving from A to C), the second cube will also have 3! = 6 pathways.
 Multiply to get 3! × 3! = 36 pathways from A to C.
 Note that we multiply the pathways from each cube (instead of adding), because the movement from A to C counts as a singlecase.

Section 5:
Other Lesson Materials 

Lecture 17 

22 pages 

Lecture 18 

00:36 

Combinations: Introduction  Welcome to Combinations. This topic will take four days to complete.
 In Day 1 we will learn how to use combinations to solve problems.
 In Day 2 we will continue our study of combinations and learn about combination notation.
 In Day 3 we will work on an assorted mix of permutations and combinations questions.
 In Day 4 we will continue to work on an assorted mix of permutations and combinations questions.


Lecture 19 

00:58 

Combinations: Summary  You have now completed Combinations. Over the past four days you have learned:
 the difference between permutations and combinations.
 how to solve combinations with or without repetitions.
 how to solve singlecase and multicase combinations.
 how to use combination notation.
 how to solve equations involving combinations.


Quiz 1 

22 questions 
Full curriculum

Excellent overview for introduction!
Videos are well organized and information is presented clearly. Work through the samples along with the instructor so the concepts sink in. You may need to review the concepts a few times to understand the material completely because there is subtly with the different types of problems encountered.