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The market after the rain
Count the complement
Sometimes the easiest way to count what you want is to count what you would leave out.
The rain has stopped. The lights are reflected in the lane, and music is drifting from a stall.
Rowan pauses to listen. Arjun, who loves sketching signs and maps, unfolds the evening programme. There are many ways to spend their visit. How many would give them the evening they want?
Rowan “I’d like to hear some music. We can fit other things around it.”
Arjun “Let’s see how many plans leave room for that.”
CKRobi “First, what will count as a different plan?”
CKSTEM Thinking CompassStuck? Find your next move.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
An evening with some music
The evening has three time slots. In each slot choose Art (A), Music (M) or Stories (S). Each activity is offered in every slot, and there is enough time to walk between them.
Repeats are allowed: you can stay for another music set or return to an activity. A plan is its three choices in time order. MMA and MAM are different plans.
How many different plans include at least one Music slot?
AArtMMusicSStories
Slot 1Choose an activity
Slot 2Choose an activity
Slot 3Choose an activity
Fill the slots from left to right. A repeat is allowed; time order distinguishes plans.Try with the CKSTEM Thinking Compass
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How many different plans include at least one Music slot?
An evening with some music
The evening has three time slots. In each slot choose Art (A), Music (M) or Stories (S). Each activity is offered in every slot, and there is enough time to walk between them.
Repeats are allowed: you can stay for another music set or return to an activity. A plan is its three choices in time order. MMA and MAM are different plans.
AArtMMusicSStories
Slot 1Choose an activity
Slot 2Choose an activity
Slot 3Choose an activity
Fill the slots from left to right. A repeat is allowed; time order distinguishes plans.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 3
How many ordered three-slot plans are there before applying the Music condition?
\(3\times3\times3=27\) plans in the whole.
Step 2 of 3
How many of those plans use only Art and Stories?
\(2\times2\times2=8\) plans have no Music.
Step 3 of 3
How many plans contain at least one Music slot?
\(27-8=19\). Compare why the two groups exhaust the same whole.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A question to help me begin
Try making a few plans. How could you avoid counting the same plan twice? Is a different group easier to describe?
Show the answer and reasoning: some music
Answer:19 plans include at least one Music slot.
Arjun “I counted plans with music first, then music second. But MMA appeared in both groups.”
CKRobi “What could you count without that difficulty?”
Rowan “Plans with no music. Every slot would be Art or Stories.”
Systematic counting
For every first choice there are three second choices. For each of those pairs there are three third choices. This gives the whole collection of plans:
\[3\times3\times3=27.\]
The plans that fail Rowan’s wish have no M anywhere. Each slot then has just two choices, A or S:
\[2\times2\times2=8.\]
AAAAASASAASSSAASASSSASSS
All eight no-music plans. Each contains only A and S.
Every ordered plan appears once. Labels as well as colours identify the two parts.
\[27-8=19.\]
Every plan either has music or has no music. None belongs to both groups, and none is left out. That is why subtraction works.
Check another way: exactly one M gives \(3\times2\times2=12\) plans: choose the M position, then A or S for each other slot. Exactly two M gives \(3\times2=6\): choose the non-M position, then A or S for it. MMM gives 1. Their total is \(12+6+1=19\). We avoid sorting those music cases when we count the simpler opposite.
Count the complement
The complement is everything in your chosen whole that does not meet the condition. Here, “no music” is the complement of “at least one music slot.”
Choose the whole carefully. Count it, count the exact unwanted part once, then subtract. Use this tool when the opposite is easier to count. If you are stuck, making just a few examples can reveal a duplicate or a simpler question.
Arjun notices Food tasting on a sampler pass. This pass has a different rule: try something different each time.
There are now four activities: A, M, S and Food tasting (F). Choose one in each of three ordered time slots, with no activity repeated. All four are offered in every slot, and walking time fits.
How many sampler plans include at least one M?
AArtMMusicSStoriesFFood tasting
Slot 1Choose an activity
Slot 2Choose an activity
Slot 3Choose an activity
The sampler uses three different activities; no choice may be used again.Try with the CKSTEM Thinking Compass
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How many sampler plans include at least one M?
A new pass changes the choices
Arjun notices Food tasting on a sampler pass. This pass has a different rule: try something different each time.
There are now four activities: A, M, S and Food tasting (F). Choose one in each of three ordered time slots, with no activity repeated. All four are offered in every slot, and walking time fits.
AArtMMusicSStoriesFFood tasting
Slot 1Choose an activity
Slot 2Choose an activity
Slot 3Choose an activity
The sampler uses three different activities; no choice may be used again.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 3
How many ordered three-slot sampler plans use no activity twice?
\(4\times3\times2=24\) plans.
Step 2 of 3
How many sampler plans have no Music and no repeat?
\(3\times2\times1=6\) unwanted plans.
Step 3 of 3
How many sampler plans include Music?
\(24-6=18\). The old repeat-allowed counts would use the wrong whole.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: the sampler pass
Answer:18 plans include Music under the sampler’s no-repeat rule.
Rowan “We can still count no-music plans. But the whole has changed.”
CKRobi “How do the choices change as you fill the slots?”
Arjun “Four choices, then three left, then two. Without Music, that becomes three, then two, then one.”
The unwanted sampler plans have no M and no repeated activity: six, rather than eight.
The six unwanted plans contain A, S and F once each, in every order. The remaining plans must contain M. Check: with no repeats, “at least one M” means exactly one M. Music has three possible positions; the other two positions have \(3\times2\) choices, giving \(3\times6=18\). This direct route is just as short.
The idea stayed the same. The old numbers 27 and 8 did not: a complement always belongs to a particular whole.
Two slots at the music stage
Rowan hears a new set starting and considers spending two slots at the music stage.
Again choose A, M or S in each of three ordered slots. Repeats are allowed, all activities run every slot and walking time fits.
How many plans have exactly two M slots? Choose a short way to count them.
Try with the CKSTEM Thinking Compass
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How many plans have exactly two M slots? Choose a short way to count them.
Two slots at the music stage
Rowan hears a new set starting and considers spending two slots at the music stage.
Again choose A, M or S in each of three ordered slots. Repeats are allowed, all activities run every slot and walking time fits.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 3
With exactly two Music slots, how many positions could the one non-Music slot occupy?
The non-Music slot can be first, second or third: three positions.
Step 2 of 3
How many activity choices are available for that non-Music slot?
There are two choices, A or S.
Step 3 of 3
How many plans have exactly two Music slots?
\(3\times2=6\). Direct counting is shorter here; a correct complement route remains valid.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: exactly two
Answer:6 plans. Direct counting is short: choose the non-M position, then choose A or S for it.
Rowan “There is only one slot that isn’t Music. It could be first, second or third.”
CKRobi “What remains to choose?”
Arjun “Art or Stories for that slot. Three positions, two choices each.”
\[3\times2=6.\]
AMMMAMMMAMMSMSMSMM
One non-M slot: choose its position, then A or S. These are all six results.
Check: every listed plan has exactly two M slots. Its non-M position and its A-or-S choice identify it uniquely. Complement would also work, but we would count the zero-M, one-M and three-M cases first: \(27-(8+12+1)=6\).
A good problem solver chooses a useful tool for this question. Knowing complement does not mean forcing every count through it.
Two wishes in one evening
Arjun wants time to sketch at Art as well as hear music with Rowan. They find a pass with four short visits, using the repeat-allowed programme.
Use four ordered slots, choosing A, M or S in each. All three are offered every slot, repeats are allowed and walking time fits.
How many plans contain at least one M and at least one A?
AArtMMusicSStories
Slot 1Choose an activity
Slot 2Choose an activity
Slot 3Choose an activity
Slot 4Choose an activity
This pass has four ordered slots, with the same three choices in each. Repeats are allowed.Try with the CKSTEM Thinking Compass
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How many plans contain at least one M and at least one A?
Two wishes in one evening
Arjun wants time to sketch at Art as well as hear music with Rowan. They find a pass with four short visits, using the repeat-allowed programme.
Use four ordered slots, choosing A, M or S in each. All three are offered every slot, repeats are allowed and walking time fits.
AArtMMusicSStories
Slot 1Choose an activity
Slot 2Choose an activity
Slot 3Choose an activity
Slot 4Choose an activity
This pass has four ordered slots, with the same three choices in each. Repeats are allowed.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 6
How many ordered four-slot plans use A, M or S, with repeats?
\(3\times3\times3\times3=81\) plans.
Step 2 of 6
How many four-slot plans have no Music?
\(2\times2\times2\times2=16\) plans have no M.
Step 3 of 6
How many four-slot plans have neither Music nor Art?
Only SSSS. It appears in both naive no-M and no-A lists.
Step 4 of 6
How many plans have Music but no Art?
\(16-1=15\). This second unwanted case is disjoint from no M.
Step 5 of 6
How many plans fail at least one of the two wishes?
\(16+15=31\) failures, each counted once.
Step 6 of 6
How many four-slot plans include both Music and Art?
\(81-31=50\). The number does not establish that the cases were complete; compare the full reasoning.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: both wishes
Answer:50 plans contain both Music and Art in the four slots.
Arjun “The whole has 81 plans. There are sixteen with no M and sixteen with no A. Is it 81 minus sixteen minus sixteen?”
CKRobi “How could you check what your subtraction removes?”
Rowan “SSSS! It has neither. We’d remove the same plan twice.”
Count the complement
The whole contains \(3\times3\times3\times3=81\) ordered plans. A plan fails if it has no Music or no Art, including plans with neither. Count these failures rather than sorting all the ways both wishes can be met. The no-M and no-A descriptions overlap, so adding their counts directly would remove a failure twice.
Rowan “I can keep every no-M plan in the first group. For the other failures, I’ll take only plans that have M but no A.”
CKRobi “Why can no plan be in both groups—and why can no failure be missed?”
Arjun “The first group has no M; the second has M. A failing plan outside the first must have M, so its missing wish must be Art.”
Casework
Separate the unwanted plans into complete cases that do not overlap: no M, then has M but no A. The first case has \(2\times2\times2\times2=16\) plans. The second uses only M and S, but excludes SSSS because it has no M: \(16-1=15\). This assigns every failure once.
SSSS belongs only to Case 1. Every four-slot plan missing at least one wish belongs to exactly one of these cases.
\[\begin{aligned}16+15&=31\text{ failures},\\81-31&=50\text{ wanted plans}.\end{aligned}\]
Check: the two groups contain every plan missing at least one wish, and neither contains a success. Together with the 50 wanted plans they account for the same whole of 81.
Another valid count
Correcting the overlap is valid: \(81-(16+16-1)=50\). A direct wanted count also works. Exactly one M gives \(4\times(2\times2\times2-1)=28\) plans. For exactly two M, the six possible pairs of M positions are (1,2), (1,3), (1,4), (2,3), (2,4) and (3,4); this gives \(6\times(2\times2-1)=18\), and exactly three M gives 4. In each case at least one remaining slot must be Art. The total is \(28+18+4=50\). The complement route avoids that three-case wanted count; disjoint Casework keeps its failure count honest.
Complement chooses the easier side of the partition. Casework makes that side complete without duplicates. Neither step alone excuses an overlapping subtraction.
Which pass keeps its promise?
A sign at the ticket stall catches Arjun’s eye: “At least 100 musical plans!” The two zone passes make the same offer. A rain notice underneath changes what that offer needs to mean.
Riverside lists 7 distinct activities: 2 music and 5 non-music. Lantern Court lists 6 distinct activities: 5 music and 1 non-music. Different music activities count as different choices.
Each pass promises at least 100 different plans containing some music, using the activities available after the closure announcement. Rowan and Arjun want that promise to hold whichever activity is cancelled.
Choose one zone pass now; it cannot be swapped. Then learn which activity has closed; then choose a plan. Exactly one listed activity in that zone will be cancelled for the whole evening. Any listed activity could be that one.
A plan fills three ordered slots. Every remaining activity is offered in every slot, repeats are allowed and walking time fits. A musical plan includes at least one music activity.
Which pass should they buy? Explain your choice.
CKSTEM Thinking CompassPlan and check your reasoning.
Which pass should they buy? Explain your choice.
Which pass keeps its promise?
A sign at the ticket stall catches Arjun’s eye: “At least 100 musical plans!” The two zone passes make the same offer. A rain notice underneath changes what that offer needs to mean.
Riverside lists 7 distinct activities: 2 music and 5 non-music. Lantern Court lists 6 distinct activities: 5 music and 1 non-music. Different music activities count as different choices.
Each pass promises at least 100 different plans containing some music, using the activities available after the closure announcement. Rowan and Arjun want that promise to hold whichever activity is cancelled.
Choose one zone pass now; it cannot be swapped. Then learn which activity has closed; then choose a plan. Exactly one listed activity in that zone will be cancelled for the whole evening. Any listed activity could be that one.
A plan fills three ordered slots. Every remaining activity is offered in every slot, repeats are allowed and walking time fits. A musical plan includes at least one music activity.
Each labelled card is a distinct activity. All remaining cards are available in each time slot.
Choose a zone
Learn which activity closes
Choose your three-slot plan
The zone choice comes before the announcement. The evening plan comes after it.A question to help me begin
When must the advertised plans be available? What information will the children have when they buy the pass, and when they make their evening plan?
Show the answer and reasoning: the advertised promise
Answer: Choose Lantern Court. It leaves 124 or 125 musical plans after the allowed closures, supporting the promise of at least 100. Riverside can leave only 91, so its promise is not supported.
A raised 3 is shorthand for three equal factors: \(7^3=7\times7\times7\). We will use it to keep the plan counts compact.
Arjun “Right now Riverside has \(7^3-5^3=218\) musical plans. Lantern Court has \(6^3-1^3=215\). Riverside looks better.”
CKRobi “What programme is the advertisement promising?”
Rowan “The one after the announcement. We can’t swap passes then. Let’s see what each choice could leave us.”
Those initial counts are correct, but they count a programme that will change.
Look one step ahead
Before committing to a pass, examine the programmes that could remain when the children make their evening plan. The pass with more plans now may fail the later promise. Count the options after the permitted closure, with the pass already fixed.
Arjun “A music closure and a non-music closure change the choices differently.”
CKRobi “Will those cases cover every activity that may close?”
Rowan “Yes. Every listed activity is either music or non-music. Activities within one type leave the same numbers of each type.”
Casework
Split the allowed closures into music and non-music. Every possible single cancellation belongs to exactly one case. Because order and repeats remain unchanged, all individual closures within one case leave the same plan count; we need no separate calculation for every activity label.
Count the complement
For each future programme, count all ordered three-slot plans from the remaining activities, then subtract those using only its remaining non-music activities. The unwanted side needs just one repeated-choice count, instead of separate wanted cases for the first music position.
Riverside
One music activity closes
6 activities remain; 5 non-music.
\(6^3-5^3=91\)
91 wanted plans
One non-music activity closes
6 activities remain; 4 non-music.
\(6^3-4^3=152\)
152 wanted plans
Least after the allowed closures: 91
Lantern Court
One music activity closes
5 activities remain; 1 non-music.
\(5^3-1^3=124\)
124 wanted plans
One non-music activity closes
5 activities remain; 0 non-music.
\(5^3-0^3=125\)
125 wanted plans
Least after the allowed closures: 124
The two cases cover every allowed single closure. Compare each remaining count with the advertised minimum of 100 plans.
At Riverside, a music closure leaves six activities, five of them non-music: \(6^3-5^3=91\). A non-music closure leaves six activities, four non-music: \(6^3-4^3=152\).
At Lantern Court, a music closure leaves five activities, one non-music: \(5^3-1^3=124\). A non-music closure leaves five music activities: \(5^3-0^3=125\).
Rowan “Lantern Court stays above 100 in both cases. Its promise holds.”
CKRobi “Does Riverside’s better case settle its advertisement?”
Arjun “No. A music activity is allowed to close, leaving only 91. The 152-plan case can’t make that failure disappear. I’d buy Lantern Court.”
Check: these are all permitted closure types. Lantern Court leaves at least 124 plans, and a music closure really attains 124; Riverside can fall to 91. The count compares available plans after the announcement, not one fixed plan that survives every closure. No assumption about equally likely cancellations is needed.
How the ideas shared the work
Count the complement counted musical plans in each changed whole.
Casework covered every allowed closure once, using two meaningful types.
Look one step ahead moved the comparison to the programme their fixed pass could actually leave them with.
Try these in your head before opening their answers. First decide what the whole contains and what it means to fail the condition.
A badge code
A badge has three code positions. Each position uses a digit from 0, 1, 2, 3, 4. Repeats and a zero in the first position are allowed: this is a code, not a three-digit number. Order matters; 012 and 120 differ.
All three positions allow 0–4. A zero in the first position is allowed; different orders make different codes.Try with the CKSTEM Thinking Compass
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How many codes contain at least one zero?
A badge code
A badge has three code positions. Each position uses a digit from 0, 1, 2, 3, 4. Repeats and a zero in the first position are allowed: this is a code, not a three-digit number. Order matters; 012 and 120 differ.
All three positions allow 0–4. A zero in the first position is allowed; different orders make different codes.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 3
How many three-position codes use digits 0–4, with repeats and leading zero allowed?
\(5\times5\times5=125\) codes.
Step 2 of 3
How many codes have no zero?
\(4\times4\times4=64\) codes have no zero.
Step 3 of 3
How many codes contain at least one zero?
\(125-64=61\) codes, including 000.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: badge codes
Answer:61 codes.
All codes: \(5\times5\times5=125\). Codes with no zero: \(4\times4\times4=64\). Subtract: \(125-64=61\).
Check: every code either contains zero or uses only 1–4. Leading-zero codes belong to the whole, including 000.
Two books to borrow
Eight distinct books are available: five mysteries and three science books. Choose two different books, with at least one science book. Order does not matter; reversing a pair does not make another choice.
How many pairs can you choose?
Mystery
Mystery 1Mystery 2Mystery 3Mystery 4Mystery 5
Science
Science 1Science 2Science 3
Each card is a distinct book. Choose two different books; reversing their order does not make a new pair.Try with the CKSTEM Thinking Compass
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How many pairs can you choose?
Two books to borrow
Eight distinct books are available: five mysteries and three science books. Choose two different books, with at least one science book. Order does not matter; reversing a pair does not make another choice.
Mystery
Mystery 1Mystery 2Mystery 3Mystery 4Mystery 5
Science
Science 1Science 2Science 3
Each card is a distinct book. Choose two different books; reversing their order does not make a new pair.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 3
Pair each of the eight books only with books later in a line. How many different unordered pairs are there?
\(7+6+5+4+3+2+1=28\) pairs, without reversals.
Step 2 of 3
How many pairs use only the five mysteries?
\(4+3+2+1=10\) unwanted pairs.
Step 3 of 3
How many pairs include at least one science book?
\(28-10=18\). Direct counting of 15 mixed and 3 science-only pairs is valid too.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning: book pairs
Answer:18 pairs.
Line the eight books up. Pair each book only with those after it: \(7+6+5+4+3+2+1=28\) pairs. Pairs with no science book use two mysteries: \(4+3+2+1=10\). So \(28-10=18\).
Check directly: one mystery and one science gives \(5\times3=15\). Two science books give three pairs. Total \(15+3=18\), with no reversal counted again.
A number trail
Consider the integers 1 through 30, including both ends.
How many are neither a multiple of 2 nor a multiple of 3?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
The whole includes every integer from 1 to 30, including both ends. No numbers have been selected yet.Try with the CKSTEM Thinking Compass
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How many are neither a multiple of 2 nor a multiple of 3?
A number trail
Consider the integers 1 through 30, including both ends.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
The whole includes every integer from 1 to 30, including both ends. No numbers have been selected yet.
Your answer
Enter the count requested by the original question. Your choice of method, complete partition and reasoning remain unscored.
This count fits. Compare your method with every original rule and the complete answer; the value alone does not establish complete reasoning.
Break it into steps
Step 1 of 5
How many multiples of 2 are in the inclusive list 1–30?
\(30\div2=15\).
Step 2 of 5
How many multiples of 3 are in the same list?
\(30\div3=10\).
Step 3 of 5
How many numbers occur in both multiple lists?
The shared numbers are 6, 12, 18, 24 and 30: five numbers in both lists.
Step 4 of 5
How many numbers are multiples of 2 or 3?
\(15+10-5=20\) unwanted numbers.
Step 5 of 5
How many numbers are neither a multiple of 2 nor a multiple of 3?
\(30-20=10\). Check the complete list and both original restrictions.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
The unwanted numbers are multiples of 2 or 3. There are 15 multiples of 2 and 10 of 3. The five multiples of 6 appear in both counts, so remove that duplicate count: \(15+10-5=20\). Then \(30-20=10\).
Check: the ten listed numbers meet both restrictions; every other number is in at least one unwanted list. Subtracting only the multiples of 6 would answer a different question.
Look back
Ask what counts as one outcome. Set the whole, including order and repetition rules. Describe the exact failure condition. If that side is easier, count it once and subtract. If direct counting is shorter, use it.
Arjun folds the programme. Rowan follows the sound of a new song. They have more than an evening plan: they have a way to turn a crowded question into a smaller one.
Continue the thinking: Look one step ahead asks what your present choice will leave you able to do next.
Rowan “I’d like to hear some music. We can fit other things around it.”
Arjun “Let’s see how many plans leave room for that.”
CKRobi “First, what will count as a different plan?”