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Permutations and combinations is the chapter where a confident method gives a confident wrong answer. The arithmetic is easy. The hard part is making sure each outcome is counted once and only once. This post shows where double counting comes from, how to catch it by listing a small case, and how to handle the standard JEE problem types.
What does the syllabus cover?
The JEE (Main) 2026 syllabus lists "The fundamental principle of counting, permutations and combinations; Meaning of P(n, r) and C(n, r). Simple applications." The JEE (Advanced) 2026 brochure includes permutations and combinations and uses them again in probability, under "computation of probability of events using permutations and combinations". NCERT's Class 11 chapter on permutations and combinations covers the counting principle, permutations and combinations. Check the documents for your exam year.
Why do counting answers go wrong?
Mathematics education researcher Elise Lockwood describes counting in terms of three parts: formulas, counting processes and sets of outcomes. In a paper on student reasoning about whether "order matters", she notes that students "often struggle" to decide between permutations and combinations, and that one way to describe their difficulty is that they do not connect their counting process to the outcomes they are trying to count. The same paper reports earlier work by Lockwood and Gibson in which listing outcomes, fully or in part, was positively correlated with correct answers for novice counters.
The practical lesson: before choosing a formula, write down what one outcome looks like. A committee is a set of people, with no order. A seating plan is an ordered arrangement. Then check your method on a case small enough to list.
Worked example 1: the "at least one" trap
A committee of 3 is chosen from 5 men and 4 women. How many committees have at least one woman?
A common wrong method: choose one woman (4 ways), then any 2 of the remaining 8 people (C(8, 2) = 28 ways), giving 112.
A correct method: count all committees and subtract those with no women. C(9, 3) − C(5, 3) = 84 − 10 = 74.
Check by cases: one woman and two men, 4 × 10 = 40; two women and one man, 6 × 5 = 30; three women, 4. Total 74.
Where did the extra 38 come from? The wrong method counts a committee with two women twice, once for each woman picked "first", and a committee with three women three times. That adds 30 and 2 × 4 = 8 extra counts, which is exactly 112 − 74 = 38.
Worked example 2: groups with and without labels
In how many ways can 6 people be split into 3 pairs?
Choose pairs one at a time: C(6, 2) × C(4, 2) × C(2, 2) = 15 × 6 × 1 = 90. But that counts the same three pairs in every order of picking them. The pairs have no labels, so divide by 3! = 6, giving 15.
If the pairs go to three named rooms, the order does matter and the answer is 90.
Check by listing a small case. With 4 people A, B, C, D, the formula gives C(4, 2) × C(2, 2)/2! = 3. The list is {AB, CD}, {AC, BD}, {AD, BC}, which is three.
Worked example 3: arrangements with repeats and circles
Repeated letters. The letters of BANANA have three A's and two N's, so the number of distinct arrangements is 6!/(3! × 2!) = 720/12 = 60. Dividing removes the swaps between identical letters, which do not create new words.
Round table. Six people around a round table can sit in (6 − 1)! = 120 ways, because rotating everyone one seat gives the same arrangement.
Necklace. For six different beads on a necklace, a flip also gives the same arrangement, so the count halves to 60.
Worked example 4: identical objects into boxes
How many ways can 10 identical sweets be given to 3 children, if a child may get none?
Line up the 10 sweets and add 2 dividers to split them into three groups. Each outcome is an arrangement of 10 sweets and 2 dividers in 12 positions, so the count is C(12, 2) = 66. If each child must get at least one, give one to each first, then share the remaining 7: C(9, 2) = 36.
Which mistakes cost the most marks?
Picking a "special" item first and then choosing freely.
Forgetting to divide when groups have no labels.
Treating rotations as different in circular arrangements.
Using permutations for selections, or combinations for arrangements.
Not checking a small case by listing.
When you get one wrong, record why, not just the right answer. Our post on why mistakes help you learn explains why that step matters.
A practice routine
Describe the outcome first (every problem): before any formula, write one sample outcome and whether order matters.
List a small case (twice a week): shrink five problems to sizes you can list by hand, list them, and check your formula.
Solve one problem two ways (weekly): complement and cases, or direct count and division. If the answers differ, find the double count.
Mix with probability. Many JEE probability questions are counting questions. Add five a week to your mixed sets, and track repeated errors with the error log for JEE plateaus.
When you cannot see why your count is too big, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the step where outcomes start to repeat.
Today, take the committee problem, change it to "at least two women", and solve it by complement and by cases.
