{ "format": "recallweave-deck/1", "title": "Counting principles: order and reuse", "attribution": "Original questions, explanations and transfer prompts authored for RecallWeave by estate-3dab0b9d2ce9 / optimizer_recovery with AI assistance. Scenarios are fictional. Mathematical background: OpenStax Precalculus 2e §11.5 and MIT Mathematics for Computer Science (2015), §§14.3–14.6. Source links and original worked transfer answers are in courses/counting-principles.md. No source exercises or illustrations are reproduced.", "license": "CC0-1.0 for this original course content. Referenced works retain their own terms and are not included in the deck.", "concepts": [ "Counting a complete choice", "Order and distinct roles", "Unordered selections", "Boundary and model checks" ], "items": [ { "id": "counting-product-1", "concept": "Counting a complete choice", "prerequisites": [], "prompt": "A badge uses exactly one of 4 backgrounds and exactly one of 3 symbols. Every background can be paired with every symbol. How many different badges can be made?", "options": [ "7", "12", "24", "64" ], "answer": 1, "explanation": "For each of the 4 backgrounds there are 3 permitted symbols, giving 4 × 3 = 12 complete choices. Adding 4 + 3 counts individual options, not background–symbol pairs. No extra position or ordering is specified.", "transfer": "A tag has 3 possible backgrounds. The first permits 2 symbols; each of the other two permits 4. How many tags are possible, and why is 3 × 4 incorrect?" }, { "id": "counting-product-2", "concept": "Counting a complete choice", "prerequisites": [], "prompt": "A display has 3 labeled positions: left, middle and right. Each position receives one of 4 icon types, and an icon type may be reused. How many different displays are possible?", "options": [ "12", "24", "81", "64" ], "answer": 3, "explanation": "Each position has 4 choices even after the earlier positions are filled, so the total is 4 × 4 × 4 = 4³ = 64. Reuse keeps the choices from decreasing. Changing the order can change which icon occupies a labeled position.", "transfer": "Change to 3 icon types and 4 labeled positions, still allowing reuse. Give the product and exact count." }, { "id": "counting-product-3", "concept": "Counting a complete choice", "prerequisites": [], "prompt": "Two labeled positions are filled with A or B, allowing reuse. AB means A in the first position and B in the second. Which list contains every ordered outcome exactly once?", "options": [ "AA, AB, BA, BB", "AA, AB, BB", "AB, BA", "AA, AB, BA, BB, AB" ], "answer": 0, "explanation": "AA, AB, BA and BB are the four complete outcomes. AB and BA differ because the positions are labeled. Removing repeated labels would wrongly exclude AA and BB; removing the reversed pair would wrongly ignore order.", "transfer": "Add a third labeled position. Write every ordered outcome and use the list to check the product count." }, { "id": "counting-order-1", "concept": "Order and distinct roles", "prerequisites": [ "Counting a complete choice" ], "prompt": "Six distinct volunteers are available. Assign one facilitator and one recorder; a person cannot hold both roles. How many assignments are possible?", "options": [ "12", "15", "30", "36" ], "answer": 2, "explanation": "There are 6 choices for facilitator and then 5 remaining choices for recorder, giving 6 × 5 = 30. The roles are distinct: swapping the same two people creates a different assignment. The count 15 would discard that distinction; 36 would allow a person to hold both roles.", "transfer": "If one named volunteer must be the facilitator, how many assignments remain? Explain which choice has become fixed." }, { "id": "counting-order-2", "concept": "Order and distinct roles", "prerequisites": [ "Counting a complete choice" ], "prompt": "Seven distinct illustrations are available. Fill the first, second and third positions of a straight display with different illustrations. How many displays are possible?", "options": [ "35", "210", "343", "5040" ], "answer": 1, "explanation": "The available choices are 7, then 6, then 5, so there are 210 displays. This is P(7,3). The count 35 ignores display order, 343 allows reuse, and 5040 arranges all seven illustrations instead of selecting three.", "transfer": "Keep the same seven illustrations but fill four labeled positions without reuse. Explain the extra factor and find the new total." }, { "id": "counting-order-3", "concept": "Order and distinct roles", "prerequisites": [ "Counting a complete choice" ], "prompt": "Choosing two different labels from A, B, C and D produces 12 ordered pairs. If order is ignored, which statement correctly counts the resulting two-label selections?", "options": [ "Each selection has exactly 2 ordered representations, so there are 12 ÷ 2 = 6 selections.", "Each selection has exactly 4 ordered representations, so there are 12 ÷ 4 = 3 selections.", "Ignoring order does not change the count, so there are 12 selections.", "Each selection can use either order, so there are 12 × 2 = 24 selections." ], "answer": 0, "explanation": "For a pair of distinct labels, AB and BA represent the same unordered selection, and those are its only two orders. Every selection is represented exactly twice, so dividing by 2 is justified. A division rule needs a uniform number of representations.", "transfer": "Choose three different labels from A, B, C and D. How many orders represent each unordered selection, and how does that convert 24 ordered outcomes into the unordered count?" }, { "id": "counting-unordered-1", "concept": "Unordered selections", "prerequisites": [ "Order and distinct roles" ], "prompt": "Choose exactly 3 different badge designs from 5 available designs. Only which designs are chosen matters; their order does not. How many selections are possible?", "options": [ "60", "15", "125", "10" ], "answer": 3, "explanation": "The 5 × 4 × 3 = 60 ordered choices represent each unordered selection 3! = 6 times. Therefore C(5,3) = 60 ÷ 6 = 10. This question asks for exactly three designs, with no reuse and no roles assigned to them.", "transfer": "Choose exactly 2 different designs from 6. Find the count, then explain why choosing the 4 designs to leave out gives the same count." }, { "id": "counting-unordered-2", "concept": "Unordered selections", "prerequisites": [ "Order and distinct roles" ], "prompt": "A box holds exactly 4 tokens. Each token has one of 3 types, and each type is available in unlimited quantity. Only the number of tokens of each type matters, not their positions. How many different box contents are possible?", "options": [ "81", "12", "15", "4" ], "answer": 2, "explanation": "This is an unordered selection with reuse: the nonnegative counts of the three types must sum to 4. Four item marks and two dividers encode each possibility once, so the count is C(4 + 3 − 1, 4) = C(6,4) = 15. Counting 3⁴ would distinguish positions that the question explicitly ignores.", "transfer": "Use 4 token types and exactly 3 tokens, with the same unlimited supply and ignored positions. How many contents are possible? State what the item marks and dividers represent." }, { "id": "counting-unordered-3", "concept": "Unordered selections", "prerequisites": [ "Order and distinct roles" ], "prompt": "Select exactly 3 labels from A and B, allowing reuse and ignoring order. A sorted string represents one multiset: AAB means two A labels and one B label. Which list represents every possible multiset exactly once?", "options": [ "AAA, AAB, ABB, BBB", "AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB", "AAA, BBB", "AAB, ABB" ], "answer": 0, "explanation": "The number of A labels can be 3, 2, 1 or 0, which produces four multisets. Different arrangements of AAB do not create new multisets. Both all-A and all-B contents are allowed because a type may be reused and no type is required to appear.", "transfer": "Increase the selection size to 4 while keeping types A and B. List the possible multisets in sorted form and check their count." }, { "id": "counting-boundary-1", "concept": "Boundary and model checks", "prerequisites": [ "Unordered selections" ], "prompt": "Select zero items from zero available label types. What is the number of outcomes in each of the four models: ordered with reuse, ordered without reuse, unordered with reuse, and unordered without reuse?", "options": [ "0 in all four models", "1 in all four models", "Undefined in all four models", "1 when reuse is allowed and 0 when reuse is forbidden" ], "answer": 1, "explanation": "All four models contain exactly one empty outcome: selecting nothing. No label needs to be available or reused. In this counting setting the empty product is 1, including the length-zero sequence over an empty alphabet; the result does not require filling any position.", "transfer": "Compare selecting zero items from five label types with selecting one item from zero types. Explain why one case has an empty outcome while the other is impossible." }, { "id": "counting-boundary-2", "concept": "Boundary and model checks", "prerequisites": [ "Unordered selections" ], "prompt": "There are only 2 distinct label types. Select exactly 3 labels without reusing either type and ignore order. How many valid selections exist?", "options": [ "1", "2", "3", "0" ], "answer": 3, "explanation": "At most two labels can be selected when neither of the two types may be reused. Selecting three is impossible, so the count is 0. It is a valid input with no outcomes, not the one empty outcome of selecting zero items.", "transfer": "Now allow reuse while keeping 2 label types and 3 selected labels. Find the ordered count and the unordered count, and explain why both become nonzero." }, { "id": "counting-boundary-3", "concept": "Boundary and model checks", "prerequisites": [ "Unordered selections" ], "prompt": "Two independent draws each return A or B with equal probability, and reuse is allowed. After drawing, ignore order, giving the three multisets AA, AB and BB. Which probabilities are correct?", "options": [ "AA: 1/3, AB: 1/3, BB: 1/3", "AA: 1/2, AB: 0, BB: 1/2", "AA: 1/4, AB: 1/2, BB: 1/4", "AA: 1/2, AB: 1/4, BB: 1/4" ], "answer": 2, "explanation": "The four equally likely ordered outcomes are AA, AB, BA and BB. The multiset AB has two ordered representations, while AA and BB each have one. Ignoring order groups outcomes but does not make the groups equally likely. A count of three groups alone cannot justify probability 1/3 for each.", "transfer": "Use three equally likely label types A, B and C for two independent draws with reuse. Compare the probabilities of multiset AA and multiset AB. Explain how their numbers of ordered representations determine the difference." } ] }