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Counting questions in the Indian Olympiad Qualifier in Mathematics (IOQM) are short to read and easy to get wrong. One case missed or one arrangement counted twice, and the integer you bubble is off. This post solves three combinatorics questions from the real IOQM 2024 paper, adds a pigeonhole example, and shows how to check each count.
What combinatorics does IOQM cover?
The Syllabus for Mathematical Olympiad on the IOQM site lists, for combinatorics, "basic enumeration, pigeonhole principle and its applications, recursion, elementary graph theory". The IOQM 2024 paper and its official answer key are on the HBCSE site. IOQM answers are integers from 00 to 99, as the HBCSE 2026-27 brochure states. For the basics of nPr and nCr, see our guide to permutations and combinations.
Question 2: fix the restricted position first
How many four-digit odd numbers use the digits 1, 2, 3 and 4, each exactly once?
The last digit must be odd, so it is 1 or 3: 2 choices. The other three digits fill the first three places in 3! = 6 ways. Total: 2 × 6 = 12. The official answer is 12.
The rule: fill the restricted position first, then count the free ones.
Question 9: count by direction
On the grid of points (m, n) with 0 ≤ m, n ≤ 4, a knight-move pair is two points that differ by 2 in one coordinate and 1 in the other. How many knight-move pairs are there?
Count unordered pairs by the step from the left point to the right point. A step of (2, 1): the left point needs m from 0 to 2 (3 choices) and n from 0 to 3 (4 choices), so 12 pairs. A step of (2, −1) also gives 12. Steps of (1, 2) and (1, −2) give 4 × 3 = 12 each. Total: 48. The official answer is 48.
Counting from each point instead would count every pair twice. Choosing a direction for each pair avoids that.
Question 19: colourings with no one-colour triangle
Five points, no three collinear, are joined in every pair, giving 10 segments. In how many ways can the segments be coloured red or blue so that no triangle has all three sides the same colour?
Each point has 4 segments. Suppose 3 or more of them were one colour, say red, going to X, Y and Z. Any red segment among X, Y, Z would make a red triangle, and if there were none, XYZ would be a blue triangle. The same argument works with the colours swapped, so each point has exactly 2 red and 2 blue segments.
The red segments therefore form a closed loop through all five points, a 5-cycle. (A triangle plus a separate pair is impossible, since two points cannot form a loop on their own.)
The blue segments, the complement of a 5-cycle among five points, form another 5-cycle with no triangle, so every red 5-cycle works.
The number of 5-cycles on five labelled points is 4!/2 = 12. The official answer is 12.
A pigeonhole example in IOQM style
This one is a practice question, not from the paper: what is the smallest n such that any n distinct numbers chosen from 1 to 100 must include two that add up to 101?
Make 50 boxes: {1, 100}, {2, 99}, ..., {50, 51}. Choosing 50 numbers, one from each box, avoids a sum of 101, so 50 is not enough. With 51 numbers, two must come from the same box, and they add to 101. The answer is 51.
Why is checking a count so hard?
Eizenberg and Zaslavsky (2004) interviewed 14 undergraduates who had taken at least one combinatorics course as they solved 10 counting problems. They describe verification as a major difficulty, identified 5 kinds of checking strategy, and found 2 of them more frequent and more helpful than the others. Lockwood (2013) proposed a model linking the formula a student writes, the counting process behind it, and the set of outcomes being counted. In practice, a count is easier to check when you can say exactly which set of outcomes your formula is counting.
Common mistakes and how to avoid them
Counting ordered pairs when the question wants unordered ones. Decide this before you start, and divide by 2 only when every pair has been counted exactly twice.
Missing a case. In Question 19, the argument has to rule out a point with 3 or 4 segments of one colour, not just assume 2 and 2.
Choosing boxes that do not force the result. In pigeonhole, show that one fewer object can avoid the property, as the 50-number example does.
A practice routine for IOQM combinatorics
Pick four counting questions from the HBCSE past papers page.
Before solving, write in words the set you are counting, such as "unordered pairs of grid points at a knight's move".
Solve, then check with a smaller case you can list by hand, such as a 3 × 3 grid.
Write one sentence on the idea that worked: "fix the restricted digit", "count by direction", "pair the numbers into boxes".
When a counting idea is new to you, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can read the idea before returning to the problem.
Next step: count the knight-move pairs on a 3 × 3 grid of points by listing them, then check the count with the direction method.
