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"At least two girls", "at most one defective bulb", "at least one book of each subject". These phrases decide whether a combinations answer is right, and they are where most marks go missing. The algebra of C(n, r) is not the hard part; the hard part is turning the words into a list of cases that covers everything once. This post gives a three-question routine for that, works four examples of different kinds, and shows how to check an answer two ways.
Where is this in the syllabus?
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" in Mathematics Unit 4, and the JEE (Advanced) 2026 brochure lists permutations and combinations in algebra. NCERT Class 11 Mathematics, Chapter 6 Permutations and Combinations (2026-27 reprint), works the "at least" type in Example 21 and sets more in its Miscellaneous Exercise. Check your own year's syllabus.
How do you turn the words into cases?
Ask three questions before writing any C(n, r).
What is being counted? A selection (order does not matter) or an arrangement?
What does the condition allow? Write the possible numbers of the special item as a list. "At least 2 out of a team of 5" with 4 available means 2, 3 or 4.
Which side is shorter? If the allowed list is long and the forbidden list short, count the total and subtract the forbidden cases. That is the complement.
"At most k" means 0, 1, ..., k. "Exactly k" means one case. "At least k" means k up to the maximum available. Always check the maximum: you cannot pick five girls from four.
Worked example 1: at most two defective bulbs
A box has 12 bulbs, of which 3 are defective. In how many ways can you choose 4 bulbs with at most 2 defective?
By cases (0, 1 or 2 defective, with the rest from the 9 good bulbs):
0 defective: C(9, 4) = 126.
1 defective: C(3, 1) × C(9, 3) = 3 × 84 = 252.
2 defective: C(3, 2) × C(9, 2) = 3 × 36 = 108.
Total: 126 + 252 + 108 = 486.
By complement (the only forbidden case is 3 defective):
Total: C(12, 4) = 495.
Forbidden: C(3, 3) × C(9, 1) = 9.
Answer: 495 − 9 = 486.
Both give 486. The complement needed one case instead of three.
Worked example 2: at least one of each kind
You choose 5 books from 4 different physics, 3 different chemistry and 3 different maths books. How many selections include at least one book of each subject?
Listing cases works but is long. Use inclusion-exclusion on the complement instead.
Total: C(10, 5) = 252.
No physics: choose all 5 from the other 6, C(6, 5) = 6.
No chemistry: C(7, 5) = 21. No maths: C(7, 5) = 21.
Missing two subjects at once would mean 5 books from at most 4, which is impossible, so nothing to add back.
Answer: 252 − 6 − 21 − 21 = 204.
A brute-force count of all 252 selections by computer also gives 204.
Worked example 3: identical items
A bag holds 3 identical red balls and 4 identical blue balls. How many selections contain at least one ball?
With identical balls, a selection is fixed by how many of each colour you take: 0 to 3 red (4 choices) and 0 to 4 blue (5 choices). That gives 4 × 5 = 20, and removing the empty selection leaves 19.
Compare distinct items: from 6 different books, the number of non-empty selections is 2⁶ − 1 = 63, because each book is either in or out. Mixing up these two formulas is a frequent error.
Worked example 4: two people who will not serve together
A committee of 4 is chosen from 10 people, but A and B refuse to serve together. How many committees are possible?
Total: C(10, 4) = 210.
Committees with both A and B: choose the other 2 from 8, C(8, 2) = 28.
Answer: 210 − 28 = 182.
By cases instead: neither, C(8, 4) = 70; A only, C(8, 3) = 56; B only, 56. Total 182.
Why do these problems trip students up?
Batanero, Navarro-Pelayo and Godino (1997) classified elementary counting problems into three models, selection, partition and distribution, and studied how the implicit model in a problem affected the reasoning of 720 pupils aged 14 to 15, before and after instruction. The practical point: the same C(n, r) can hide behind very different wording. Naming the model and listing the cases, as in the three questions above, keeps the wording from steering you. For double-counting traps such as "pick one woman first, then the rest", see our guide to avoiding double counting in permutations and combinations.
Which mistakes cost marks?
Reading "at most 2" as "fewer than 2". It includes 2.
Forgetting the zero case in "at most".
Listing a case that is impossible, such as 5 girls from 4.
Using 2ⁿ − 1 for identical items.
Choosing one special item first, then filling freely, which counts some selections more than once.
A practice routine
Solve NCERT Miscellaneous Exercise 6, Question 3 (exactly, at least and at most 3 girls) by cases and check parts (ii) and (iii) by complement.
Do Question 7, at least 3 from each part, and list the allowed splits first.
Redo example 2 with 4 books chosen instead of 5. Now one double-missing case is possible: all 4 physics books. You should get 126.
Each week, mix five selection problems with five from other chapters; Dunlosky and colleagues (2013) rated practice testing and spaced practice as high utility.
When an answer key gives a number but not the case list, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, solve example 1 again for "at least 2 defective" and check that your two answers add up to 495 plus the "exactly 2" count.
