Search constrained combinations
Reduce a multi-variable search with equations, bounds and divisibility, then verify every surviving case.
Section 26-2.3 · 2026–2027 · 2.3 · Deduce and compare outcomes
Original StudyForA scenarios and formative model answers; no Cambridge marks, predicted grades or certified mastery. Alternative reasoned answers may be valid. Full-length unseen papers, a wider mixed-problem bank and calibrated specialist teacher marking remain outstanding.
Before you begin
- Read tables and instructions in English
- Use arithmetic, ratios and whole-number constraints
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Before calculating, what should you identify?
What you will learn
- Search a larger finite possibility space efficiently
- Prove the list of solutions is complete
Reduce the number of free choices
A total-count equation can eliminate one variable. A weighted-total equation can then constrain the others. Use non-negative bounds before listing cases; divisibility and parity can remove many cases immediately. Keep a record of each rejected case so a completeness claim has evidence.
Select 6 packages: small cost 2, medium 4, large 7 credits, spending exactly 24. Let counts be s,m,l. From s+m+l=6 and 2 s+4 m+7 l=24, obtain 2 m+5 l=12. The possible l values are 0,1,2; only l=0,m=6,s=0 or l=2,m=1,s=3 work.
Explain all constraints, including leftovers
A derived equation can permit a case with a negative leftover s; reject it even if the cost equation balances. For ordered arrangements distinguish a combination from a sequence. Do not multiply by permutations unless different orders represent different outcomes.
A well-presented search states its starting bounds, the reduction and the final verification. This remains readable when later parts change the cost, count or available stock.
Sources for this lesson
Pause and explain
Why can l be at most two in 2 m+5 l=12 with non-negative m,l?
Worked example
List every package selection meeting the six-package and twenty-four-credit rules.
Show the worked solution
- Eliminate s to get 2 m+5 l=12.
- l=0 gives m=6; l=1 gives non-integer m; l=2 gives m=1.
- Recover s and check both totals.
Answer The selections (s,m,l) are (0,6,0) and (3,1,2). At l≥3 the derived cost exceeds twelve, so no further non-negative case is possible.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Guided: Verify the (3,1,2) package solution.
Give me a hint
Check count and weighted price.
Compare my reasoning
- Count is 3+1+2=6.
- Cost is 3×2+1×4+2×7=24.
- All counts are non-negative integers.
The selection meets all conditions: six packages, twenty-four credits, and valid whole counts. Verification is needed even after algebraic elimination.
Look for these in your work
- Checks count
- Checks price
- Checks domain
Independent: Select 5 packages at the same prices, spending exactly 20. Find all count triples.
Give me a hint
Eliminate s using the count total.
Compare my reasoning
- The cost equation becomes 2 m+5 l=10.
- l=0 gives m=5,s=0; l=1 gives fractional m.
- l=2 gives m=0,s=3; higher l is impossible.
The only triples are (0,5,0) and (3,0,2). Their prices are 20 and 6+14=20, respectively, and each has five packages.
Look for these in your work
- Derives the reduced constraint
- Enumerates the bounded cases
- Checks both totals
Application: In the original six-package problem, at most four medium packages are in stock. Which solution survives?
Give me a hint
Apply the new stock limit to the complete original list.
Compare my reasoning
- The original cases are (0,6,0) and (3,1,2).
- Six medium packages exceeds stock.
- One medium package is within stock.
Only (3,1,2) survives. The stock condition removes the all-medium selection; it cannot introduce a new solution because the original count and price rules remain unchanged.
Look for these in your work
- Uses the complete list
- Applies the stock limit
- Explains why no new case appears
Common mistakes
- Discarding the count constraint
- Keeping negative leftovers
- Claiming completeness without bounds
What can you explain now?
Why can l be at most two in 2 m+5 l=12 with non-negative m,l?
Compare with the explanation
Three large packages already contribute fifteen, exceeding twelve
The bound follows from the constraint, not from an arbitrary search length.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Why can l be at most two in 2 m+5 l=12 with non-negative m,l?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.