A problem-solving strategy

Make an organized list

Every possibility. Exactly once. A reason to trust the answer.

Mina and Li Wei compare seed-picture cards at a school garden worktable, with blue CKRobo, pink CKRobi and green CKSprout thinking alongside them.
At the garden worktable, the team compares possible seed displays.

Mina and her friend Li Wei are preparing a garden display. Li Wei loves asking whether an idea really works. Their coach, CKSprout—Sprout for short—helps them think through the choices.

Visitors will choose a pair of seed varieties. The children want one picture card for every different pair, so every visitor can see all the choices.

Mina “I’ve made a few cards. But how will we know when we have them all?”

Li Wei “And how will we know we haven’t counted the same choice twice?”

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?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

01

The missing display cards

There are four seed varieties: Basil, Bean, Marigold and Sunflower. Each card shows two different varieties. A pair is the same choice whichever variety is named first: Basil with Bean is the same pair as Bean with Basil.

Which different pairs should the display show—and how can the team be certain none is missing?

Mina’s first cards are a mixed-up start:

1BasilBean
2BasilSunflower
3BeanBasil
4MarigoldSunflower
A first pile of four cards. Which choices are different?

Li Wei “These two cards name the same pair in a different order. Four cards don’t necessarily mean four different choices.”

CKSprout “What would make the missing choices easier to notice?”

A growing pile of guesses is hard to check. The team needs a rule for what comes next.

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Give the search an order

Make an organized list: choose an order and follow it, so you find every allowed choice once.

Use it to find all the ways, test choices against a clue, or check whether there is just one answer.

A neat list of guesses might still miss a choice. Your system should explain why nothing is missed and why nothing is repeated.

02

Finish one group before moving on

Show the answer and reasoning

Answer: There are 6 different pairs, listed below.

Mina “Let’s keep the varieties in this order: Basil, Bean, Marigold, Sunflower.”

Li Wei “For each pair, we’ll name the earlier variety first. Then a reversed pair won’t get another card.”

  1. Start with Basil. Pair it with each later variety: Bean, Marigold and Sunflower.
  2. Move to Bean. Basil–Bean is already included. The new partners are Marigold and Sunflower.
  3. Move to Marigold. Its only later partner is Sunflower.
  4. Stop at Sunflower. It has no later partner. Every earlier partnership has already been considered.

Start with Basil3 new pairs

1BasilBean
2BasilMarigold
3BasilSunflower

Start with Bean2 new pairs

4BeanMarigold
5BeanSunflower

Start with Marigold1 new pair

6MarigoldSunflower
The earlier variety fixes the group; every later partner appears once.

\[3+2+1=6\text{ pairs}\]

The grid shows the same system. A variety cannot pair with itself here. A reversed pair is the same choice, so only one of its two positions is used.

Ba BasilBe BeanM MarigoldS Sunflower
Use only the numbered cells: one place for every different pair.
WithBaBeMS
Ba—123
BeSeen—45
MSeenSeen—6
SSeenSeenSeen—
— same variety, not allowed. “Seen” marks a reversed pair already counted. Numbers match the six cards above.

Why can we trust this list?

Every allowed pair has one variety earlier in the fixed order. Its card belongs to that variety’s group, so every pair has a place. We included every later partner in that group. No pair can belong to two different earlier-variety groups, so no pair is counted twice.

CKSprout “What did the system reveal about the first pile?”

Mina “It repeated Basil–Bean. It missed Basil–Marigold, Bean–Marigold and Bean–Sunflower.”

The six cards are the answer. The grouping rule is the reason we know the answer is complete.

Systematic counting

Split the possibilities into groups that cover every allowed choice and do not overlap. Then add the group counts.

Another way to check the count

The plant groups contain three, two and one new pairs. We can also count two-position choices first: four choices for the first variety and three different choices for the second.

\[4\times3=12\]

That counts Basil–Bean and Bean–Basil separately. For these seed pairs, every pair has been counted exactly twice, so:

\[12\div2=6\]

The division is justified by the two counted orders. It is not a rule to apply to every choice problem.

Change the question: does order matter?

Li Wei now designs a display with a left pot and a right pot. For each pot he can choose Basil, Bean, Marigold or Sunflower, but the two pots must contain different varieties. Two displays count as different if either pot contains a different variety.

How many different left-and-right displays can he make?

Mina “Can we count these the same way as the seed pairs?”

CKSprout “What makes two displays different here?”

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Show the answer and reasoning

Answer: There are 12 different left-and-right displays.

Left potBasil
Right potBean
Left potBean
Right potBasil
Two different displays: left and right are distinct positions.

Li Wei “We can still choose each left-hand variety in turn. This time, all three other varieties are allowed on the right.”

Mina “So Basil on the left and Bean on the right stays different from Bean on the left and Basil on the right. The two positions do different jobs.”

\[4\times3=12\text{ displays}\]

Before listing or counting, ask “What makes two choices different in this problem?” The objects stayed the same; the question changed.

03

A gift bag with more moving parts

The team makes gift bags holding six seed packets. Only Basil, Bean and Sunflower packets are used, and every bag must contain at least one of each. Two bags are the same mix when their three packet counts match; moving packets around inside a bag does not create a new mix.

How many different mixes are possible?

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A question to help me begin

Could you keep one packet count fixed while you change another?

Show the answer and reasoning

Answer: There are 10 different packet mixes.

Mina “If we change all three counts at once, it will be easy to skip a mix.”

Li Wei “Let’s fix the Basil count first. For each Basil count, increase Bean one at a time. Sunflower gets the packets left.”

There can be at most four Basil packets: one Bean and one Sunflower must still fit. With one Basil packet, Bean can be one, two, three or four. With two Basil packets, Bean can be one, two or three—and so on.

BasilBeanSunflower

Fix Basil at 14 mixes

1 Basil, 1 Bean, 4 Sunflower

1 Basil, 2 Bean, 3 Sunflower

1 Basil, 3 Bean, 2 Sunflower

1 Basil, 4 Bean, 1 Sunflower

Fix Basil at 23 mixes

2 Basil, 1 Bean, 3 Sunflower

2 Basil, 2 Bean, 2 Sunflower

2 Basil, 3 Bean, 1 Sunflower

Fix Basil at 32 mixes

3 Basil, 1 Bean, 2 Sunflower

3 Basil, 2 Bean, 1 Sunflower

Fix Basil at 41 mix

4 Basil, 1 Bean, 1 Sunflower

Every strip has six packets. Read the named counts as well as the colours.

\[4+3+2+1=10\]

Every allowed bag has exactly one Basil count, then exactly one Bean count. Those two counts force its Sunflower count. Every permitted first count and second count is included, so every mix appears once.

Check a row against every rule.

For example, two Basil, one Bean and three Sunflower packets give \(2+1+3=6\). All three counts are positive. Reordering those six packets produces the same mix.

A list is only as good as its rules

The first cards needed two different varieties. The bags could repeat a variety, but needed all three types. When the rules change, check what counts as one choice again.

Sorting your first guesses can help you find a system. Then ask: does my system reach every allowed choice?

If your list gets very long, look for a rule that lets you skip steps safely. You still need a reason that the skipped cases follow your rule.

A small change to think through

The team can stamp a label with two symbols chosen from Leaf, Sun, Drop and Star. The same symbol may be used twice. Swapping the two symbols counts as the same label design.

How many different label designs are there?

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A question to help me begin

Use the fixed symbol order. When starting a group, should its first symbol be allowed to pair with itself?

Show the answer and reasoning

Answer: There are 10 different label designs.

The Leaf group has four partners, including Leaf. The Sun group has three new partners, the Drop group two, and the Star group one.

Leaf Leaf
Leaf Sun
Leaf Drop
Leaf Star
Sun Sun
Sun Drop
Sun Star
Drop Drop
Drop Star
Star Star
Equal symbols are allowed; reversed versions of different symbols are not repeated.

\[4+3+2+1=10\]

Each pair is recorded with its earlier symbol first, allowing equality. That includes every permitted pair without recording its reversal again.

Why not count all sixteen ordered choices and divide by two? A pair such as Leaf–Leaf has only one order, while Leaf–Sun has two. Not every design was counted twice.

04

The missing packing note

The packing note for a seedling delivery is missing. Every tray was either small or large and was full on arrival: a small tray held 3 seedlings and a large tray held 5 seedlings. There was at least one tray of each size.

The team first donated half the original seedlings to another class. Then they planted 7 seedlings in their display. They had 14 seedlings left.

Donation and planting moved individual seedlings, not necessarily full trays. All seedlings stayed whole, and no others were added or removed.

Find every possible mix of small and large trays that could have arrived. Do the clues tell us which mix actually arrived?

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A question to help me begin

Which facts describe the start, and which describe the finish? What must a complete delivery explain?

Show the answer and reasoning

Answer: There are two possible mixes: 9 small trays and 3 large trays, or 4 small trays and 6 large trays. The clues do not tell us which of these actually arrived.

Mina “Planting seven happened last. Before that, there were twenty-one seedlings.”

CKSprout “Which moment does that twenty-one describe?”

Li Wei “After the donation. Half remained, so the original delivery had twice as many: forty-two.”

Work backwards

Start with the known finish and recover earlier stages in reverse order. Put back the planted seedlings before undoing the earlier half-donation:

\[\begin{aligned}14+7&=21\\21\times2&=42\end{aligned}\]

Reasoning backwards · return to the delivery

14Seedlings left
+ 7Put back the planted seedlings
21After donation
× 2Recover the original whole
42On arrival
Follow the arrows from the known finish back to arrival. Undo the last event before the earlier donation.

The original total is now known. It still does not tell us how that total was divided between the two tray sizes.

Li Wei “I’ll increase the large-tray count one at a time. Each count leaves a fixed number of seedlings for small trays.”

CKSprout “How will you know where to stop?”

Mina “Both sizes were there. Eight large trays and even one small tray would already hold forty-three. I only need to try one through seven large trays.”

\[8\times5+3=43>42\]

Mina “One large tray leaves thirty-seven seedlings for small trays. Thirty-seven cannot fill whole trays of three.”

CKSprout “What has that trial told you?”

Li Wei “It rules out that large count, not every mix. I’ll continue through seven.”

Check every allowed large-tray count, from 1 to 7.
Large traysSeedlings in themSeedlings for small traysDoes the remainder fit?
1537Cannot fill whole small trays
21032Cannot fill whole small trays
315279 whole small trays
42022Cannot fill whole small trays
52517Cannot fill whole small trays
630124 whole small trays
7357Cannot fill whole small trays
Every matching row describes a complete mix. The two matching rows both need to be retained.

Mina “Three large trays leave \(42-15=27\) seedlings. Those fill \(27\div3=9\) small trays.”

Li Wei “Six large trays leave \(42-30=12\). Those fill \(12\div3=4\) small trays. We have two matching rows.”

CKSprout “What shows that your list is complete?”

Mina “We checked every allowed large count. Each fixes one small count, so no mix is missed or repeated. Finding the first matching row would have missed the second one.”

Replay both deliveries.

\[\begin{aligned}9\times3+3\times5&=42\\4\times3+6\times5&=42\end{aligned}\]

For either mix, donate half: \(42-21=21\). Plant seven: \(21-7=14\). Both use full trays of both sizes, and every seedling count is whole.

CKSprout “What can the clues tell us about which mix actually arrived?”

Li Wei “Both deliveries end with fourteen seedlings. We cannot choose between them without another clue.”

How the strategies shared the work

Work backwards recovered the original delivery once. Make an organized list then found the complete set of tray mixes. Knowing a total did not force one packing.

05

A bigger CIC delivery

For their Canadian Innovation Contests (CIC) garden display, the team receives a separate delivery of 40 full trays, all either small or large. Small trays hold 3 seedlings, large trays hold 5, and there is at least one tray of each size.

  1. First, they donate one quarter of the original seedlings to the school garden.
  2. Then they plant 18 seedlings in their CIC display.
  3. They have 108 seedlings left.

Donation and planting use individual seedlings, not necessarily whole trays. All seedlings stay whole, and no others are added or removed.

How many small trays and how many large trays arrived? Could a different mix fit every clue?

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A question to help me begin

What can you settle before choosing a tray mix? How could you keep track of all the allowed choices?

Show the answer and reasoning

Answer: There were 16 small trays and 24 large trays. No different mix fits every clue.

Recover the total before choosing a tray mix

Mina “I’ll put back the eighteen planted seedlings first. That makes one hundred and twenty-six after donation.”

CKSprout “How is that amount connected to the original delivery?”

Li Wei “It is the three quarters left. One quarter is forty-two, so four quarters make one hundred and sixty-eight.”

\[\begin{aligned}108+18&=126\\126\div3&=42\\42\times4&=168\end{aligned}\]

Reasoning backwards · return to the delivery

108Seedlings left
+ 18Put back the planted seedlings
126After donation
÷ 3, then × 4Recover the original whole
168On arrival
Follow the arrows from the known finish back to arrival. Undo the last event before the earlier donation.

Begin a complete search in a fixed order

Li Wei “Both sizes must be there. I’ll start with one large and thirty-nine small, then increase the large count one at a time. The last possible large count is thirty-nine.”

CKSprout “What fixes the small count in each row?”

Mina “There are forty trays altogether. One large leaves thirty-nine small; two large leave thirty-eight small; and so on.”

Every allowed large count from 1 to 39 has exactly one partner. That gives a complete order without missing or repeating a mix.

Mina “The first three totals are one hundred and twenty-two, one hundred and twenty-four, and one hundred and twenty-six. The total rises by two.”

CKSprout “What would show that the change keeps working?”

Li Wei “Each step replaces one tray of three with one tray of five. There are still forty trays, and every replacement adds two seedlings.”

Check the first rows, then explain the change before skipping rows.
Large traysSmall traysTotal seedlings
139122
238124
337126
436128
The intervening large counts are 5 through 38. We will explain the rule that covers them.
391198
The tray counts always add to forty. Each permitted large count appears once in the complete order.

Look for a pattern

Notice a relationship, explain why it continues, then use it to reach a case you have not calculated yet. Here the replacement explains the change in every permitted row, not just the first few.

Small: 3 seedlings
Replace one tray+2 seedlings
Large: 5 seedlings
One tray replaces one tray. The extra two seedlings explain every step, not just the first few.

\[5-3=2\]

Mina “Our legal rows run from one hundred and twenty-two to one hundred and ninety-eight seedlings, so one hundred and sixty-eight lies between them. The first row has one large tray. We need forty-six more seedlings from there.”

CKSprout “Where does that put you in your order?”

Li Wei “Each step adds two, so it takes twenty-three more steps. I started at one large tray: one plus twenty-three is twenty-four large trays.”

\[\begin{aligned}168-122&=46\\46\div2&=23\text{ more steps}\\1+23&=24\text{ large trays}\\40-24&=16\text{ small trays}\end{aligned}\]

Twenty-four is within the allowed large counts. Every step strictly increases the total by two, so no other row can give the same 168 seedlings. The explained rule lets us jump within a complete list instead of calculating all 39 totals.

Check every original clue.

There are \(16+24=40\) full trays, including both sizes. The small trays hold \(16\times3=48\) seedlings; the large trays hold \(24\times5=120\). Together:

\[48+120=168\]

Donate a whole quarter, then plant eighteen:

\[\begin{aligned}168\div4&=42\\168-42-18&=108\end{aligned}\]

Every seedling count is whole, and the finish is exactly right.

How the strategies shared the work

  1. Make an organized list gave every permitted tray mix one position, with a clear first legal row.
  2. Work backwards recovered the delivery total from the finish, reversing the last event first.
  3. Look for a pattern justified the constant change and reached the needed row without calculating every intervening total.

A complete order can reveal a useful relationship. Explain why it continues before using it to skip rows.

Your turn

Try a new setting. Decide what counts as different and choose your own system before opening an answer.

A small museum donation

You want to put exactly 25 cents in a donation box, using only 5-cent and 10-cent coins. You have enough of both types. You may use none of one type. Swapping coins of the same value, or changing their order, does not make a new way.

What are all the different coin-count mixes, and how many are there?

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Show the answer and reasoning

Answer: There are 3 mixes: five 5-cent coins; three 5-cent coins and one 10-cent coin; or one 5-cent coin and two 10-cent coins.

List the 10-cent count in order: zero, one, two. Three would already be 30 cents, too much. The money still needed is 25, 15 or 5 cents, which fixes the 5-cent count at five, three or one. Every allowed 10-cent count is included once.

Check all three.

\(5\times5=25\), \(3\times5+10=25\), and \(5+2\times10=25\). Zero of a type is allowed; different coin orders were not counted again.

Two-scoop choices

An ice-cream shop has mango, vanilla, chocolate and strawberry. Every serving has exactly two scoops. Both scoops may have the same flavour.

How many cup choices and how many cone choices are there?

  1. In a cup, only the flavours matter. Mango–vanilla and vanilla–mango count as the same choice. How many cup choices are there?
  2. On a cone, the top and bottom flavours matter. Mango on top with vanilla below is different from vanilla on top with mango below. How many cone choices are there?
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Show the answers and reasoning

Answer: There are 10 cup choices and 16 cone choices.

For cups, keep the flavour order above. Start with mango and use it with mango, vanilla, chocolate or strawberry: four choices. Then vanilla with vanilla, chocolate or strawberry: three new choices. Then chocolate with chocolate or strawberry: two. Finally strawberry with itself: one.

\[4+3+2+1=10\]

Every cup has exactly one place in this order. For cones, fix each top flavour in turn. Every top has four bottom choices, so \(4\times4=16\).

Check what changed.

The six cup choices with two different flavours each give two cone orders. The four same-flavour cups each give one: \(6\times2+4=16\). This also explains why \(16\div2\) would not count the cups correctly.

The music club’s boxes

A music club packs 10 full boxes of egg shakers. Small boxes hold 4 shakers; large boxes hold 7. There are 52 shakers altogether, with at least one box of each size.

How many small boxes and how many large boxes are there? Is there another mix that fits?

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Show the answer and reasoning

Answer: There is just one mix: 6 small boxes and 4 large boxes.

For a reference count, imagine all ten boxes are small (this is not an allowed final mix): \(10\times4=40\) shakers. Each swap to a large box adds \(7-4=3\). The club has \(52-40=12\) extra shakers, so it needs \(12\div3=4\) large boxes. The other six are small.

Check both totals.

\(6+4=10\) boxes and \(6\times4+4\times7=52\) shakers. Both sizes are present. Each extra large box changes the total by three, so a second mix cannot give the same total.

Look back

Mina and Li Wei began with a messy pile of cards. A fixed order revealed both a repeated pair and missing choices. Later, the same habit helped them compare packet mixes. Their complete tray list showed two possible deliveries; more clues would be needed to identify which one actually arrived. In a larger delivery mystery, an explained pattern let them extend the list and jump to the right row.

What counts as different?
What rule tells me what comes next?
Why does every allowed choice appear exactly once?
Do the clues identify one answer, or leave several possible?

Keep those questions with you when choosing teams, arranging objects, finding number combinations or checking possible explanations. The aim is not the longest list. It is a search you can explain and trust.

Continue with Look for a pattern, where a rule can take you beyond a short list.

Another person may choose a different order. Compare the reasons their method is complete; different routes can reach the same reliable answer.

CKSTEM Thinking Framework

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