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How is the total number of outcomes calculated?","back":"Multiply the options: $a \\times b$.","mcqOptions":[{"text":"Multiply the options: $a \\times b$.","correct":true,"explanation":"For two independent stages, each of the $a$ first choices can be paired with each of the $b$ second choices, so the total is found by multiplication."},{"text":"Add the options: $a + b$.","correct":false,"explanation":"Adding counts the total number of individual choices available, not the number of two-stage outcomes. The product rule uses $a \\times b$ because every first choice can pair with every second choice."},{"text":"Subtract the options: $a - b$.","correct":false,"explanation":"Subtraction would compare the two numbers of choices rather than count all possible pairings. For a two-stage task, the total number of outcomes is $a \\times b$."},{"text":"Divide the options: $a \\div b$.","correct":false,"explanation":"Division shares one number of choices by the other, which does not count combinations. The correct product rule is $a \\times b$, one pairing for each first-stage and second-stage choice."}]},{"id":"629867f8-b040-413d-886c-4da14bb7b740","cardType":"cloze","points":null,"front":"When choosing one item from each of several separate groups, you must [ ] the number of choices in each group.","back":"When choosing one item from each of several separate groups, you must multiply the number of choices in each group.","mcqOptions":[{"text":"multiply","correct":true,"explanation":"Separate groups of choices form combinations by pairing each choice from one group with each choice from the others, so the numbers of choices must be multiplied."},{"text":"add","correct":false,"explanation":"Adding is tempting because it finds the total number of items, but it does not count combinations. Choosing one item from each separate group requires multiplying the group sizes."},{"text":"subtract","correct":false,"explanation":"Subtraction compares two quantities instead of counting all possible selections. For one choice from each group, use multiplication so every possible pairing is included."},{"text":"divide","correct":false,"explanation":"Division would split or share choices, not count the combinations formed across groups. The product rule says to multiply the number of choices in each group."}]},{"id":"5501d987-59fd-486d-9b0a-6a78197c7394","cardType":"basic","points":1,"front":"A lunch offer has 2 wraps and 3 drinks. How many possible lunch combinations are there?","back":"$$2 \\times 3 = 6$.","mcqOptions":[{"text":"$$2 \\times 3 = 6$.","correct":true,"explanation":"Each of the 2 wraps can be matched with each of the 3 drinks, giving $2 \\times 3 = 6$ possible lunch combinations."},{"text":"$$2 + 3 = 5$.","correct":false,"explanation":"Adding counts the number of menu items, not the number of wrap-and-drink pairings. The product rule gives $2 \\times 3 = 6$ combinations."},{"text":"$$3 - 2 = 1$.","correct":false,"explanation":"Subtracting only compares the numbers of wraps and drinks, so it cannot count possible lunches. Each wrap pairs with each drink, giving $2 \\times 3 = 6$."},{"text":"$$2^3 = 8$.","correct":false,"explanation":"Using a power treats the choice as if there were 3 repeated stages with 2 options each. There are just 2 wraps and 3 drinks, so $2 \\times 3 = 6$."}]},{"id":"9444a642-bcee-443d-8870-6d59cd8ad06d","cardType":"basic","points":2,"front":"Why is addition ($11 + 13$) incorrect for finding the number of boy-and-girl pairs from 11 boys and 13 girls?","back":"Addition gives the total number of students, not the number of possible combinations.","mcqOptions":[{"text":"Addition gives the total number of students, not the number of possible combinations.","correct":true,"explanation":"The sum $11 + 13$ counts students separately. A boy-and-girl pair needs one choice from each group, so the number of possible pairs comes from multiplying."},{"text":"Addition gives the number of boys only, not the number of possible combinations.","correct":false,"explanation":"The sum $11 + 13$ does not give the number of boys only; it gives the total number of students. The real issue is that pairs require multiplication, not addition."},{"text":"Addition gives the number of girls only, not the number of possible combinations.","correct":false,"explanation":"The sum $11 + 13$ does not give the number of girls only; it gives all students together. To count boy-and-girl pairs, multiply the two group sizes."},{"text":"Addition gives the number in one pair, not the number of possible combinations.","correct":false,"explanation":"One pair contains 2 students, not $11 + 13$ students. Adding gives the total number of students, while multiplying counts every possible boy-and-girl pairing."}]},{"id":"f1860d99-b64b-4b85-bd39-573e6bce09b0","cardType":"cloze","points":null,"front":"In counting problems, a calculated number of choices ($x$) must always be a [ ].","back":"In counting problems, a calculated number of choices ($x$) must always be a whole number.","mcqOptions":[{"text":"whole number","correct":true,"explanation":"A number of choices counts actual possible options, so it cannot be a part of an option or a negative amount. It must be a whole number."},{"text":"proper fraction","correct":false,"explanation":"A proper fraction would mean less than one complete choice, which is impossible when counting outcomes. A calculated number of choices must be a whole number."},{"text":"negative number","correct":false,"explanation":"A negative value would mean fewer than no choices, which cannot represent a count of outcomes. Counts in these problems must be whole numbers."},{"text":"prime number","correct":false,"explanation":"Some numbers of choices are prime, but many valid counts are not, such as 6 or 12. The necessary condition is being a whole number."}]},{"id":"ad71cd34-b883-4147-82e4-d46b5ac73111","cardType":"basic","points":1,"front":"A restaurant offers 4 starters, 6 mains, and 3 desserts. How many different three-course meals are possible?","back":"$$4 \\times 6 \\times 3 = 72$.","mcqOptions":[{"text":"$$4 \\times 6 \\times 3 = 72$.","correct":true,"explanation":"A three-course meal is formed by choosing one starter, one main, and one dessert. The product rule gives $4 \\times 6 \\times 3 = 72$ meals."},{"text":"$$4 + 6 + 3 = 13$.","correct":false,"explanation":"Adding counts the total number of dishes on the menu, not the number of meals. Each starter can combine with each main and dessert, so $4 \\times 6 \\times 3 = 72$."},{"text":"$$4 \\times 6 + 3 = 27$.","correct":false,"explanation":"This multiplies starters and mains but then only adds desserts, missing that each dessert can go with every starter-main pair. The correct calculation is $4 \\times 6 \\times 3 = 72$."},{"text":"$$4 \\times 3 \\times 6 = 72$.","correct":false,"explanation":"Although this gives the same value, it does not match the stated order of starters, mains, and desserts. The intended product-rule calculation is $4 \\times 6 \\times 3 = 72$."}]},{"id":"dcc8abb7-6f2e-45ce-83ee-f05e55d1dd72","cardType":"basic","points":1,"front":"What condition must be met to use the product rule directly for multiple stages?","back":"Every option at one stage must be allowed to go with every option at the next stage.","mcqOptions":[{"text":"Every option at one stage must be allowed to go with every option at the next stage.","correct":true,"explanation":"The product rule works directly only when there are no restrictions between stages. Then each choice at one stage can be paired with every choice at the next."},{"text":"Every option at one stage must be different to go with every option at the next stage.","correct":false,"explanation":"Options being different is not the key condition; restrictions are. The product rule applies directly when every option at one stage is allowed with every option at the next."},{"text":"Every option at one stage must be equally likely to go with every option at the next stage.","correct":false,"explanation":"Equally likely choices matter in probability, not in simply counting outcomes. For the product rule, the requirement is that every cross-stage combination is allowed."},{"text":"Every option at one stage must be chosen to go with every option at the next stage.","correct":false,"explanation":"You do not choose every option; you choose one option at each stage. Direct multiplication is valid when each possible first choice is allowed with each possible next choice."}]},{"id":"5dd14437-5077-4b3c-afd3-65890e16ec2a","cardType":"basic","points":2,"front":"A shop has 8 sandwiches and $x$ drinks. Sam claims there are 90 meal deal outcomes. Why is Sam wrong?","back":"

$90 \\div 8 = 11.25$
The number of drinks must be a whole number.

","mcqOptions":[{"text":"

$90 \\div 8 = 11.25$
The number of drinks must be a whole number.

","correct":true,"explanation":"If there are 8 sandwiches and $x$ drinks, the outcomes are $8x$. Solving $8x = 90$ gives $x = 11.25$, which cannot be a count of drinks."},{"text":"

$90 \\times 8 = 720$
The number of drinks must be a whole number.

","correct":false,"explanation":"Multiplying by 8 is the reverse of the required step. Since $8x = 90$, divide by 8 to get $x = 11.25$, not a whole number."},{"text":"

$90 \\div 8 = 11.25$
The number of drinks can be a decimal.

","correct":false,"explanation":"The division is right, but the conclusion is not. A shop cannot have 11.25 drink choices, so the number of drinks must be a whole number."},{"text":"

$90 \\div 8 = 12$
The number of drinks must be a whole number.

","correct":false,"explanation":"This rounds the division instead of calculating it exactly. $90 \\div 8 = 11.25$, and that non-whole result is why Sam’s claim cannot be correct."}]},{"id":"4003d549-23fc-4b55-88f2-99bc5874207c","cardType":"cloze","points":null,"front":"Choosing \"without replacement\" means the number of options for the next stage [ ].","back":"Choosing \"without replacement\" means the number of options for the next stage decreases by 1.","mcqOptions":[{"text":"decreases by 1","correct":true,"explanation":"Once an item has been selected without replacement, it is unavailable at the next stage. Therefore, the number of available options falls by exactly one."},{"text":"stays the same","correct":false,"explanation":"The number stays the same only when selection is with replacement. Without replacement, the chosen item is unavailable, so the number of options decreases by 1."},{"text":"increases by 1","correct":false,"explanation":"This reverses the effect of removing an item from the available choices. Without replacement, one option is used up, so the number decreases by 1."},{"text":"decreases by 2","correct":false,"explanation":"This assumes two options are removed after one choice. Only the selected item becomes unavailable, so the number of options decreases by 1."}]},{"id":"eaa74b2f-bdac-453e-9254-b3f62a78731a","cardType":"basic","points":1,"front":"Two cards are dealt from a 52-card deck to two named people. How many outcomes are possible?","back":"$$52 \\times 51 = 2652$.","mcqOptions":[{"text":"$$52 \\times 51 = 2652$.","correct":true,"explanation":"The first named person can receive any of 52 cards, while the second can receive any of the remaining 51. The product rule gives $52 \\times 51 = 2652$."},{"text":"$$52 \\times 52 = 2704$.","correct":false,"explanation":"Using $52$ choices at both stages incorrectly allows the same card to be dealt twice. Without replacement, the correct calculation is $52 \\times 51 = 2652$."},{"text":"$$\\frac{52 \\times 51}{2} = 1326$.","correct":false,"explanation":"Dividing by 2 treats the recipients as interchangeable. Because the two people are named, swapping their cards creates a different outcome, giving $52 \\times 51 = 2652$."},{"text":"$$51 \\times 51 = 2601$.","correct":false,"explanation":"Using $51$ choices for both stages omits one possible first card. There are 52 initial choices and then 51 remaining choices, giving $2652$."}]},{"id":"129a4be2-a62f-4817-ad3a-2ec1f1fadb17","cardType":"basic","points":2,"front":"Why is the second stage $51$ instead of $52$ when dealing two cards from a deck?","back":"One card has already been removed and cannot be chosen again.","mcqOptions":[{"text":"One card has already been removed and cannot be chosen again.","correct":true,"explanation":"Dealing is without replacement, so the first card is no longer available for the second person. This leaves exactly 51 possible cards at the second stage."},{"text":"One card has already been removed but can be chosen again.","correct":false,"explanation":"A removed card cannot immediately be selected again when dealing without replacement. Consequently, the second stage has 51 available cards rather than 52."},{"text":"One person has already been named and cannot be chosen again.","correct":false,"explanation":"The number falls because one physical card has been removed, not because the recipients are named. After the first deal, only 51 cards remain available."},{"text":"One ordering has already been counted and cannot be counted again.","correct":false,"explanation":"Dividing for reversed arrangements applies to unordered pairs, not to the number of second-stage choices. The second stage is 51 because one card has been removed."}]},{"id":"c19d0b24-ae36-4644-b539-a721a274f93f","cardType":"basic","points":1,"front":"When calculating unordered pairs from $n$ items, what is the formula?","back":"$$$\\frac{n(n-1)}{2}$$","mcqOptions":[{"text":"$$$\\frac{n(n-1)}{2}$$","correct":true,"explanation":"There are $n(n-1)$ ordered selections of two distinct items. Each unordered pair appears in both possible orders, so dividing by 2 gives $\\frac{n(n-1)}{2}$."},{"text":"$$$n(n-1)$$","correct":false,"explanation":"This counts every unordered pair twice because it retains both possible orders. Dividing by 2 is required, so the correct formula is $\\frac{n(n-1)}{2}$."},{"text":"$$$\\frac{n(n+1)}{2}$$","correct":false,"explanation":"The second factor must be $n-1$ because the first selected item cannot be selected again. Unordered pairs are therefore counted by $\\frac{n(n-1)}{2}$."},{"text":"$$$\\frac{(n-1)(n-2)}{2}$$","correct":false,"explanation":"This incorrectly removes two choices before either item has been selected. There are $n$ first choices and $n-1$ second choices, then division by 2."}]},{"id":"4ca8946e-80e5-42d1-913f-dbd61fe86167","cardType":"cloze","points":null,"front":"For handshakes or matches, we [ ] because each pair is counted twice (e.g., A-B and B-A).","back":"For handshakes or matches, we divide by 2 because each pair is counted twice (e.g., A-B and B-A).","mcqOptions":[{"text":"divide by 2","correct":true,"explanation":"The product $n(n-1)$ counts A-B and B-A separately. Since these represent the same handshake or match, dividing by 2 counts each pair once."},{"text":"multiply by 2","correct":false,"explanation":"Multiplying by 2 would count each pair four times after both orders are already included. The duplicate orderings must instead be removed by dividing by 2."},{"text":"subtract 2","correct":false,"explanation":"Subtracting 2 removes only two outcomes, rather than one duplicate for every pair. Because every pair is counted twice, the total must be divided by 2."},{"text":"divide by 4","correct":false,"explanation":"Dividing by 4 removes too many outcomes because each pair appears only twice, not four times. The correct adjustment is therefore to divide by 2."}]},{"id":"fb45b2ad-ff16-427b-ad02-e631a0fbd177","cardType":"basic","points":2,"front":"28 students are in a class. Two are chosen for the same prize. How many possible pairs are there?","back":"$$$\\frac{28 \\times 27}{2} = 378$$","mcqOptions":[{"text":"$$$\\frac{28 \\times 27}{2} = 378$$","correct":true,"explanation":"There are $28 \\times 27$ ordered selections of two different students. Because the prize is the same, each pair is counted in both orders, giving $378$ after division by 2."},{"text":"$$$28 \\times 27 = 756$$","correct":false,"explanation":"This counts each pair twice, once in each selection order. Since both students receive the same prize, order is irrelevant and division by 2 gives $378$."},{"text":"$$$28 \\times 28 = 784$$","correct":false,"explanation":"Squaring 28 incorrectly allows the same student to occupy both selections and also treats order as significant. Distinct unordered pairs give $\\frac{28 \\times 27}{2}=378$."},{"text":"$$$\\frac{27 \\times 26}{2} = 351$$","correct":false,"explanation":"This starts with 27 rather than all 28 possible first students, creating an off-by-one error. The correct calculation is $\\frac{28 \\times 27}{2}=378$."}]},{"id":"3c65a4d6-c903-47fb-8d53-de4d54b9655d","cardType":"basic","points":2,"front":"9 teams play each other once in a league. How many games are played?","back":"$$$\\frac{9 \\times 8}{2} = 36$$","mcqOptions":[{"text":"$$$\\frac{9 \\times 8}{2} = 36$$","correct":true,"explanation":"Each of the 9 teams can face 8 others, producing 72 ordered listings. Every game appears from both teams’ perspectives, so division by 2 gives 36."},{"text":"$$$9 \\times 8 = 72$$","correct":false,"explanation":"This counts every game twice, once from each team’s perspective. Since a game between A and B is the same as B against A, the answer is 36."},{"text":"$$$9 \\times 9 = 81$$","correct":false,"explanation":"Using $9^2$ incorrectly includes each team playing itself and counts games between different teams twice. The correct calculation is $\\frac{9 \\times 8}{2}=36$."},{"text":"$$$\\frac{8 \\times 7}{2} = 28$$","correct":false,"explanation":"This calculation uses only 8 teams, omitting one team and all its fixtures. With 9 teams, the correct total is $\\frac{9 \\times 8}{2}=36$."}]},{"id":"241f91d4-0b52-439b-b07f-bc50eda84fd6","cardType":"cloze","points":null,"front":"If two people are chosen for different roles, you [ ] divide by 2.","back":"If two people are chosen for different roles, you do not divide by 2.","mcqOptions":[{"text":"do not","correct":true,"explanation":"Different roles make order significant: choosing A then B assigns them differently from choosing B then A. Both assignments must remain, so division by 2 is inappropriate."},{"text":"always","correct":false,"explanation":"Dividing by 2 would wrongly merge two distinct role assignments. When the roles differ, swapping the chosen people changes the outcome, so you do not divide by 2."},{"text":"must","correct":false,"explanation":"Division by 2 is required for identical roles or unordered pairs, not different roles. Different assignments remain separate outcomes, so you do not divide by 2."},{"text":"should","correct":false,"explanation":"This confuses different roles with an unordered selection. Since reversing the two people changes who holds each role, both orders count and you do not divide by 2."}]},{"id":"87ba5ccf-ccb6-4063-9ed2-38517af4e617","cardType":"basic","points":2,"front":"Contrast an option with an outcome.","back":"

Option: A choice at a single stage.
Outcome: A complete result after all stages.

","mcqOptions":[{"text":"

Option: A choice at a single stage.
Outcome: A complete result after all stages.

","correct":true,"explanation":"An option is one available choice within a stage, whereas an outcome is the complete result formed by combining choices from every stage."},{"text":"

Option: A complete result after all stages.
Outcome: A choice at a single stage.

","correct":false,"explanation":"This reverses the two definitions. An option is a choice at one stage, while an outcome is a complete result after all stages have been completed."},{"text":"

Option: The number of choices at every stage.
Outcome: The number of stages in total.

","correct":false,"explanation":"This confuses choices and results with numerical counts. An option is one choice at a single stage, while an outcome is one complete result after all stages."},{"text":"

Option: A complete list of possible results.
Outcome: A method used to count choices.

","correct":false,"explanation":"This confuses individual choices and results with lists and methods. An option is a choice at one stage, while an outcome is a complete result after all stages."}]},{"id":"ca35884a-b804-4789-b2a2-83a60aaa9c27","cardType":"cloze","points":null,"front":"In product rule questions, the word [ ] usually indicates that you should multiply the choices.","back":"In product rule questions, the word \"and\" usually indicates that you should multiply the choices.","mcqOptions":[{"text":"\"and\"","correct":true,"explanation":"The word \"and\" links successive stages that must all occur. The product rule therefore multiplies the number of choices available at each stage."},{"text":"\"or\"","correct":false,"explanation":"The word \"or\" usually indicates alternative cases whose numbers of outcomes should be added. The word \"and\" links successive stages, so their choices are multiplied."},{"text":"\"not\"","correct":false,"explanation":"The word \"not\" indicates an exclusion or restriction, rather than the product rule. The word \"and\" usually signals that choices from successive stages should be multiplied."},{"text":"\"only\"","correct":false,"explanation":"The word \"only\" places a restriction on the permitted outcomes and does not itself signal multiplication. The usual product-rule indicator is the word \"and\"."}]},{"id":"5be842dd-5c66-4304-8ed9-f44ce44f37f5","cardType":"basic","points":1,"front":"When should you divide by 2 in a counting problem?","back":"When making pairs, matches, or handshakes where the order does not matter.","mcqOptions":[{"text":"When making pairs, matches, or handshakes where the order does not matter.","correct":true,"explanation":"Choosing A with B and choosing B with A describe the same unordered pair. Basic multiplication counts each pair twice, so dividing by 2 corrects the duplication."},{"text":"When making lists, codes, or passwords where the order does matter.","correct":false,"explanation":"Ordered lists, codes, and passwords treat different arrangements as different outcomes, so dividing by 2 would wrongly remove valid results. Division by 2 applies to unordered pairs counted twice."},{"text":"When making pairs, matches, or handshakes where the order does matter.","correct":false,"explanation":"If order matters, AB and BA are distinct outcomes and should both remain counted. Divide by 2 only for pairs, matches, or handshakes where order does not matter."},{"text":"When making choices, selections, or decisions where the stages are independent.","correct":false,"explanation":"Independent successive stages normally require multiplying their numbers of options, not dividing by 2. Division is needed for unordered pairs when multiplication has counted each pair twice."}]},{"id":"0ae5b4b0-22d1-4fe9-9fa1-a912250dc197","cardType":"basic","points":1,"front":"Identify the 3 stages in choosing a three-course meal.","back":"

Choosing a starter.
Choosing a main.
Choosing a dessert.

","mcqOptions":[{"text":"

Choosing a starter.
Choosing a main.
Choosing a dessert.

","correct":true,"explanation":"A three-course meal is formed through three successive choices: one starter, one main, and one dessert. The product rule multiplies the available options at these stages."},{"text":"

Choosing a starter.
Choosing a drink.
Choosing a dessert.

","correct":false,"explanation":"A drink is not one of the three courses specified in a three-course meal. The stages are choosing a starter, choosing a main, and choosing a dessert."},{"text":"

Choosing a starter.
Choosing a main.
Choosing a drink.

","correct":false,"explanation":"Replacing dessert with a drink misidentifies the final course. The three stages are choosing a starter, choosing a main, and choosing a dessert."},{"text":"

Choosing a main.
Choosing a dessert.
Choosing a drink.

","correct":false,"explanation":"This omits the starter and incorrectly includes a drink as a course. The stages are choosing a starter, choosing a main, and choosing a dessert."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"The Product Rule for Counting","title":"The Product Rule for Counting Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"04672075-d371-49f3-ade4-f5af53c49689","walled":"$L39"}],"$L3a","$L3b"]