Permutation & Combination - Misconception Buster
Permutation and combination problems are among the most common sources of exam errors for Filipino students. These misconceptions can cost you 10-15 marks in major exams like UPCAT, CSE, and NMAT. Understanding where students typically go wrong is crucial because these errors follow predictable patterns. This guide identifies the most critical mistakes and provides trap questions to test your understanding.
Common misconceptions
Permutation and combination are the same thing - order doesn't matter in either case
Exam impact
This misconception leads to using the wrong formula entirely, resulting in completely incorrect answers. Can cost 3-5 marks per question.
The reality
Permutation cares about ORDER (nPr = n!/(n-r)!), combination does NOT care about order (nCr = n!/[r!(n-r)!]). The extra r! in combination formula specifically removes the ordering effect.
Trap question
Question
In how many ways can a director choose 3 actors from 10 to star in a movie where all three have equal importance?
Since all three actors have equal importance, the order of selection doesn't matter. This is a combination problem, not permutation.
Wrong answer
720 (using 10P3 = 10×9×8)
Correct answer
120 (using 10C3 = 10!/(3!×7!) = 10×9×8/(3×2×1))
Correct vs incorrect approach
Correct approach
Using nCr for committee selection because the order of selection doesn't create different committees
Incorrect approach
Using nPr when selecting a committee of 3 from 10 people, thinking arrangement matters
Why students believe it
Students see both formulas involving factorials and n!/r! patterns, making them think they're interchangeable. The word 'arrangement' appears in both contexts, causing confusion.
Keywords always tell you which formula to use - 'arrange' means permutation, 'choose' means combination
Exam impact
Students blindly follow keyword rules and choose wrong formulas, especially in word problems that deliberately use misleading language.
The reality
Keywords can be misleading. The context and whether order matters in the final result determines the formula. You can 'arrange' a committee (combination) or 'choose' positions (permutation).
Trap question
Question
A teacher wants to arrange 5 students into a study group. In how many ways can this be done?
The word 'arrange' here means forming a group, not positioning them. Since all 5 students must be in the group, there's only one way.
Wrong answer
120 (using 5P5 = 5! because of the word 'arrange')
Correct answer
1 (there's only one way to form a group of 5 from 5 students)
Correct vs incorrect approach
Correct approach
Analyzing whether the arrangement creates meaningfully different outcomes - committee membership vs seating positions
Incorrect approach
Seeing 'arrange students in a committee' and automatically using permutation
Why students believe it
Students memorize keyword lists without understanding context. Teachers often give simplified rules that don't account for all scenarios.
In combination problems, you can multiply by r! at the end to get permutation
Exam impact
Students solve combination problems correctly but then multiply by r! thinking they need permutation, getting wrong answers.
The reality
This mathematical relationship exists but doesn't mean the problems are equivalent. Converting combination to permutation changes the problem's meaning entirely.
Trap question
Question
From 8 books, you need to select 4 for a reading list. How many different reading lists are possible?
A reading list is just a selection of books - the order doesn't matter. Multiplying by 4! would be counting different arrangements of the same books, which isn't what's asked.
Wrong answer
1680 (calculating 8C4 = 70, then multiplying by 4! = 24)
Correct answer
70 (using 8C4 = 8!/(4!×4!) directly)
Correct vs incorrect approach
Correct approach
Understanding that selection problems need combination only - the multiplication changes it to a different problem
Incorrect approach
Solving 'select 3 students from 10' as 10C3 = 120, then multiplying by 3! = 6 to get 720
Why students believe it
Students notice that nPr = nCr × r! and think this conversion always works in reverse for checking answers.
Factorial calculation: 0! = 0 and undefined factorials equal zero
Exam impact
Wrong factorial values lead to completely incorrect calculations, especially in edge cases like nCn or nC0.
The reality
0! = 1 by definition (important for combination formulas), and factorials of negative numbers are undefined, not zero. This affects formulas when n = r.
Trap question
Question
In how many ways can you choose all 6 items from a group of 6 items?
There's exactly one way to choose all items from a group - take everything. The formula works because 0! = 1 by mathematical definition.
Wrong answer
0 (thinking 6C6 = 6!/(6!×0!) and 0! = 0)
Correct answer
1 (since 6C6 = 6!/(6!×1) = 1, because 0! = 1)
Correct vs incorrect approach
Correct approach
Calculating 5C5 = 5!/(5!×0!) = 5!/(5!×1) = 1
Incorrect approach
Calculating 5C5 = 5!/(5!×0!) = 5!/(5!×0) = undefined or 0
Why students believe it
Students think 0! should logically equal 0 (zero times anything is zero), and when they can't calculate factorials like (-1)!, they assume it's 0.
When objects are repeated or identical, you always divide by the factorial of repetitions
Exam impact
Students unnecessarily complicate problems by dividing by factorials when dealing with any identical objects, getting smaller incorrect answers.
The reality
Division by repetition factorials only applies when you're arranging objects and the identical ones create indistinguishable arrangements. For selection problems, identical objects don't affect the count the same way.
Trap question
Question
How many ways can you select 4 marbles from a bag containing 3 red, 3 blue, and 3 green marbles?
This is a selection problem with repetition allowed, not an arrangement problem. We count distinct ways to distribute 4 selections among 3 colors.
Wrong answer
84 divided by 3! = 14 (applying repetition division incorrectly)
Correct answer
15 (using systematic counting: 4-0-0, 3-1-0, 2-2-0, 2-1-1 distributions)
Correct vs incorrect approach
Correct approach
Recognizing this as selecting from distinct groups: choose from {A,B,C} with repetition allowed
Incorrect approach
Choosing 3 letters from AABBCC and dividing by 2!×2!×2! for the repeated letters
Why students believe it
Students learn the formula for arrangements with repetition and apply it everywhere, even when the repetition isn't relevant to the question.
nCr = nC(n-r) is just a coincidence and doesn't mean anything important
Exam impact
Students miss opportunities to simplify calculations, potentially making computational errors with large factorials.
The reality
This property reflects that choosing r objects is equivalent to choosing which (n-r) objects to leave out. It's useful for simplifying calculations when r > n/2.
Trap question
Question
What's the most efficient way to calculate 50C47?
The symmetry property allows us to use the smaller number (3 instead of 47) for much easier calculation.
Wrong answer
Calculate 50!/(47!×3!) directly
Correct answer
Use 50C47 = 50C3 = 50×49×48/(3×2×1) = 19,600
Correct vs incorrect approach
Correct approach
Using 20C17 = 20C3 = 20×19×18/(3×2×1) = 1140
Incorrect approach
Calculating 20C17 as 20!/(17!×3!) with large factorial operations
Why students believe it
Students see this symmetry property mentioned but don't understand its logical basis or practical applications for simplifying calculations.
Circular arrangements always use (n-1)! regardless of the problem context
Exam impact
Students apply circular formulas incorrectly, often getting answers that are too small by factor of 2 or applying the formula to non-circular problems.
The reality
Circular arrangements use (n-1)! only when rotations are considered identical but reflections are different. If reflections are also identical, use (n-1)!/2. Context determines which formula applies.
Trap question
Question
In how many ways can 6 people sit around a circular table if clockwise and counterclockwise arrangements are considered the same?
When reflections (clockwise vs counterclockwise) are considered identical, we must divide by 2 to avoid double counting.
Wrong answer
120 (using (6-1)! = 5! = 120)
Correct answer
60 (using (6-1)!/2 = 5!/2 = 120/2 = 60)
Correct vs incorrect approach
Correct approach
Using (n-1)!/2 when both rotations and reflections are considered identical
Incorrect approach
Using (n-1)! for arranging people around a circular table when the problem considers clockwise and counterclockwise as the same
Why students believe it
Students memorize the circular permutation formula without understanding when and why it applies, using it for any problem mentioning circles or round tables.
In 'at least' or 'at most' problems, you can just use the basic formula with the minimum or maximum value
Exam impact
These problems are common in major exams and students get completely wrong answers by oversimplifying, missing most of the cases.
The reality
'At least' and 'at most' problems require summing multiple cases or using complementary counting. There's no single direct formula for these compound conditions.
Trap question
Question
A committee of 5 people must be formed from 8 men and 6 women, with at least 2 women. How many ways can this be done?
'At least 2 women' means 2, 3, 4, or 5 women. We must calculate and sum all valid cases, not just the minimum.
Wrong answer
6C2 × 8C3 = 15 × 56 = 840 (only considering exactly 2 women)
Correct answer
2002 (sum of cases: 2W+3M, 3W+2M, 4W+1M, 5W+0M = 840 + 896 + 240 + 6 = 1982, or use complement)
Correct vs incorrect approach
Correct approach
Calculating C(5,2) + C(5,3) + C(5,4) + C(5,5) or using complement: 2^5 - C(5,0) - C(5,1)
Incorrect approach
For 'at least 2 heads in 5 coin flips', just calculating C(5,2)
Why students believe it
Students want to avoid complex calculations and hope that using the boundary value will give the right answer directly.
Probability in permutation/combination problems is just the favorable outcomes divided by any reasonable total
Exam impact
Students get probability answers that are impossible (>1) or drastically wrong because they mix counting methods.
The reality
In probability problems involving permutation/combination, the sample space and favorable outcomes must be counted using the same method - both with order or both without order.
Trap question
Question
What's the probability that a randomly selected 3-person committee from 10 people includes exactly 2 specific individuals?
We need exactly 2 specific people plus 1 from the remaining 8. Total ways to form committee is C(10,3), favorable ways is C(8,1).
Wrong answer
2/10 = 0.2 (thinking it's just the ratio of specific people)
Correct answer
C(8,1)/C(10,3) = 8/120 = 1/15 ≈ 0.067
Correct vs incorrect approach
Correct approach
Using C(4,2)/C(52,2) or P(4,2)/P(52,2), keeping counting method consistent
Incorrect approach
Finding probability of drawing 2 aces from 52 cards using C(4,2) as favorable but 52×51 as total
Why students believe it
Students know the basic probability formula but don't ensure that the numerator and denominator count the same type of outcomes (both ordered or both unordered).
Consecutive number problems always use simple arithmetic progression formulas
Exam impact
Students miss the connection between consecutive arrangements and permutation concepts, using wrong mathematical approach.
The reality
Some consecutive problems involve arrangements or selections of consecutive numbers, requiring permutation/combination formulas. The word 'consecutive' doesn't automatically mean use AP formulas.
Trap question
Question
In how many ways can you arrange the consecutive numbers 1, 2, 3, 4 in a row such that 1 is not in the first position?
This is a conditional permutation problem. We find total arrangements minus restricted arrangements, not simple subtraction.
Wrong answer
Just subtract 1 from total positions: 4-1=3
Correct answer
18 (Total arrangements 4! = 24, minus arrangements starting with 1 which is 3! = 6, so 24-6=18)
Correct vs incorrect approach
Correct approach
Recognizing arrangement of consecutive numbers still follows permutation rules
Incorrect approach
For arranging consecutive numbers 1,2,3,4,5, using sum formula instead of 5!
Why students believe it
Students see patterns in consecutive numbers and try to use AP formulas or simple addition rather than recognizing permutation/combination applications.
Quick self-check
If a problem asks you to 'select' items, it's always a combination problem
Answer
False
Context matters more than keywords. You could select people for specific positions (permutation) or select items for a group where order doesn't matter (combination).
0! equals 1, not 0
Answer
True
By mathematical definition, 0! = 1. This is crucial for combination formulas when r = n or r = 0.
nC5 equals nC(n-5) for any valid value of n
Answer
True
This symmetry property holds because choosing r items is equivalent to choosing which (n-r) items to leave out.
Circular arrangements of n objects always give (n-1)! possibilities
Answer
False
This is true only when rotations are identical but reflections are different. When reflections are also identical, use (n-1)!/2.
In probability problems, you can mix permutation counting in numerator with combination counting in denominator
Answer
False
Both numerator and denominator must use the same counting method to ensure proper probability calculation.
The formula nPr can be converted to nCr by dividing by r!
Answer
True
nCr = nPr/r! because combination removes the ordering counted in permutation.
'At least 3' means you only need to calculate the case with exactly 3
Answer
False
'At least 3' means 3 OR more, so you must sum all cases from 3 up to the maximum, or use complementary counting.
When dealing with identical objects, you always divide by the factorial of repetitions
Answer
False
This only applies to arrangement problems where identical objects create indistinguishable arrangements, not to all problems with identical objects.
Diagrams
Permutation vs Combination Decision Tree
Visual guide to distinguish when order matters (permutation) versus when order doesn't matter (combination)
Problem-Solving Flowchart for P&C Word Problems
Step-by-step flowchart to avoid keyword traps and properly analyze P&C problems
Common Factorial Misconceptions
Shows how factorial misconceptions propagate through P&C formulas
At Least Problems: Wrong vs Right Approach
Demonstrates the difference between incomplete and complete solutions for 'at least' problems
In summary
The key to avoiding permutation and combination mistakes is understanding the fundamental question: 'Does order matter in the final result?' Don't rely solely on keywords - analyze the context. Remember that 0! = 1, use symmetry properties for efficient calculation, and always sum all cases for 'at least/at most' problems. When solving probability problems, keep your counting method consistent throughout. These misconceptions cost students significant marks in major Philippine exams, but understanding them will give you a clear advantage.