1. Define the counted outcome in words
State in words what one counted outcome is before choosing any formula.
State in words what one counted outcome is. This prevents jumping to a formula before the model is clear.
JEE JEE Main and JEE Advanced · Mathematics
Model a counting problem correctly before choosing a formula by deciding whether order matters, whether repetition is allowed, whether objects are distinguishable, and what restrictions apply.
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In short
Permutation and combination formulas are outputs of a counting model, not the starting point. First decide what is being counted.
If order matters, use an arrangement model. If order does not matter, use a selection model. Then decide whether repetition is allowed, whether any objects are indistinguishable, and whether restrictions split the problem into cases.
The official JEE documents define content scope. They do not publish chapter weightage, so none is asserted here.
Verified against the current JEE Main 2026 syllabus and JEE Advanced 2026 syllabus.
Sources: JEE Main 2026 syllabus and JEE Advanced 2026 syllabus, both linked in the sources section below.
State in words what one counted outcome is before choosing any formula.
State in words what one counted outcome is. This prevents jumping to a formula before the model is clear.
Order determines whether the model is a permutation or a combination.
Decide whether changing order creates a different outcome. If it does, use a permutation-based model; if not, use a combination-based model.
Repetition changes the formula from factorial-based to power-based counting.
Decide whether an option may be reused. Do not assume repetition is allowed simply because a formula would be convenient.
Interchanging identical objects does not create a new outcome.
Identify identical objects before counting arrangements. Treating identical objects as distinguishable overcounts the result.
Restrictions can split a problem into disjoint cases or be handled through complements.
Apply restrictions by cases, complement or position fixing, then only select nPr, nCr, factorial division or a product of choices once the model is settled.
Total outcomes equal the product of the choice counts at each stage.
Count of outcomes for a staged process with successive independent choices.
Use when — Sequential choices.
Common trap — Multiplying mutually exclusive alternatives.
Add the counts of cases that do not overlap.
Add counts of disjoint cases.
Use when — Either/or cases.
Common trap — Double-counting overlap.
n factorial equals n times n minus one and so on down to one; zero factorial equals one.
Number of ways to arrange n distinct objects.
Use when — Arrange n distinct objects.
Common trap — Undefined negative factorial.
n P r equals n factorial divided by n minus r factorial.
Number of ordered arrangements of r objects chosen from n distinct objects, no repetition.
Use when — Arrange r from n.
Common trap — Using when order does not matter.
n C r equals n factorial divided by r factorial times n minus r factorial.
Number of unordered selections of r objects from n distinct objects, no repetition.
Use when — Choose r from n.
Common trap — Counting order repeatedly.
n P r equals n C r times r factorial.
Relation between ordered and unordered counts.
Use when — Convert selection then arrangement.
Common trap — Forgetting r!.
n to the power r.
Count of ordered sequences of r labelled positions, each with n available choices, repetition allowed.
Use when — Codes or sequences under stated rules.
Common trap — Using when restrictions change by position.
n factorial divided by the product of the factorials of each identical group's size.
Distinct arrangements of n objects with repeated identical groups.
Use when — Repeated identical objects.
Common trap — Dividing for objects that are actually distinguishable.
Answer: 30
Five letters total. L appears twice, E appears twice, V appears once. Order matters, but the L copies and E copies are indistinguishable.
If all five positions contained labelled objects there would be 5! arrangements. Each visible arrangement is counted 2! times for the L copies and 2! times for the E copies.
5!/(2!2!) = 120/4 = 30.
Validity check: the result is smaller than 5!=120, as it must be because identical-letter swaps do not create new outcomes.
Choosing a formula before defining the counted outcome.
Decision / selection errorWhy it happens
Without a defined outcome, it is unclear whether order, repetition or distinguishability apply.
How it is corrected
State in words what one counted outcome is before selecting a formula.
Using permutation when order does not matter.
Decision / selection errorWhy it happens
This overcounts outcomes that are actually the same selection.
How it is corrected
Confirm whether order changes the outcome before choosing nPr or nCr.
Using combination when order does matter.
Decision / selection errorWhy it happens
This undercounts outcomes that should be treated as distinct.
How it is corrected
Confirm whether order changes the outcome before choosing nPr or nCr.
Assuming repetition is allowed because the formula is convenient.
Needs reviewWhy it happens
Repetition must be justified by the problem statement, not assumed.
How it is corrected
Check the problem wording explicitly for repetition allowance.
Treating identical objects as distinguishable.
Execution errorWhy it happens
This overcounts arrangements by the number of internal permutations of the identical group.
How it is corrected
Identify identical objects and divide by their internal factorial.
Adding overlapping cases.
Execution errorWhy it happens
The addition principle requires disjoint cases.
How it is corrected
Check for intersections between cases before adding their counts.
Multiplying mutually exclusive alternatives.
Decision / selection errorWhy it happens
Multiplication applies to sequential independent choices, not either/or alternatives.
How it is corrected
Use the addition principle for either/or alternatives instead.
Double-counting when restrictions create overlapping cases.
Needs reviewWhy it happens
Restriction cases that overlap will be counted more than once if added directly.
How it is corrected
Verify restriction cases are disjoint, or use complement reasoning instead.
Official syllabus scope verified against NTA JEE Main 2026 and JEE Advanced 2026 documents. No weightage, frequency or PYQ-count claim is made.
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