Combinatorics in 1 Shot ISI–CMI Maths 2026 by Vivek Bansal Sir

Combinatorics is one of the most powerful and frequently tested areas in ISI and CMI entrance examinations. It blends counting techniques, number theory, and logical reasoning into complex problem-solving scenarios. 

Here, Vivek Bansal Sir simplifies selected Previous Year Questions (PYQs) from ISI/CMI exams to help aspirants understand advanced strategies like the Pigeonhole Principle, Stars and Bars method, modular arithmetic, and permutation restrictions.

This approach is especially useful for students preparing for ISI, CMI, JEE Advanced, and Olympiads, where deep conceptual clarity is essential for solving unconventional and high-difficulty problems.

Also Read:

• Geometry & Coordinate Geometry Problem Solving for ISI, CMI, and IOQM
• Number Theory in 1 Shot ISI–CMI Maths Mastery 

Problem 1: ISI UGCB 2023 Combinatorics Question

Let n1, …, n51 be 51 distinct natural numbers, each with exactly 2023 positive integer factors. No prime factor greater than 11 divides any ni. We must prove at least one perfect cube exists among these numbers. Hint: 2023 = 7 * 17 * 17.

For a number N = p1^a * p2^b * p3^c, the number of positive factors is (a + 1)(b + 1)(c + 1). Allowed prime factors are {2, 3, 5, 7, 11}.

Since 2023 = 7 * 17 * 17, numbers with 2023 factors must have exponent structures (X+1) multiplying to 2023. This leads to four types:

1. Type 1: p^2022

2. Type 2: p1^6 * p2^288

3. Type 3: p1^16 * p2^118

4. Type 4: p1^16 * p2^16 * p3^6
   (Memory Tip: This problem heavily uses the Pigeonhole Principle (PHP).)

Counting and Perfect Cube Identification

Using the 5 allowed primes:

• Type 1 (p^2022): 5 ways (e.g., 2^2022). Perfect cube (2022 is divisible by 3).

• Type 2 (p1^6 * p2^288): 5C2 * 2! = 20 ways. Perfect cube (6, 288 are divisible by 3).

• Type 3 (p1^16 * p2^118): 5C2 * 2! = 20 ways. (Not a perfect cube; 16, 118 are not divisible by 3).

• Type 4 (p1^16 * p2^16 * p3^6): 5C1 * 4C2 = 30 ways. (Not a perfect cube; 16 is not divisible by 3).
  This results in 25 Perfect Cubes (Type 1 + Type 2) and 50 Non-Perfect Cubes (Type 3 + Type 4), totaling 75 unique numbers.

Application of Pigeonhole Principle

We select 51 distinct numbers from the 75 possible. By the Pigeonhole Principle, to avoid selecting a perfect cube, one would first pick all 50 non-perfect cubes. The 51st number must be a perfect cube. Therefore, at least one perfect cube exists among the chosen numbers.

Problem 2: ISI UGB 2022 Board Problem

A board has 2 rows, n columns (2n cells), filled with 0 or 1. Each row/column sum parity depends on the number of 1s (even sum for even 1s, odd sum for odd 1s).

Part 1: Every Row and Column Sum is Even

To ensure even column sums, the second row must be identical to the first row. This automatically makes the second row sum even if the first row sum is even. The problem reduces to filling the first row with an even number of 1s. The number of ways is given by nC0 + nC2 + nC4 + … = 2^(n-1).

Part 2: Every Row and Column Sum is Odd

To ensure odd column sums, the second row must be the complement of the first row.

• If n is Odd: If the first row has 'k' (odd) ones, the second row will have (n-k) ones. Since n (odd) - k (odd) = even, the second row sum becomes even, contradicting the problem's requirement. Result: 0 ways.

• If n is Even: If the first row has 'k' (odd) ones, the second row will have (n-k) ones. Since n (even) - k (odd) = odd, both row sums will be odd. The problem reduces to filling the first row with an odd number of 1s. The number of ways is given by nC1 + nC3 + nC5 + … = 2^(n-1).

Problem 3: ISI UGA 2025 Four-Tuple Problem

Find the number of different four-tuples (A, B, C, D) of positive integers where A * B * C * D = 999.

Prime Factorization: 999 = 3^3 * 37^1.

Each A, B, C, D must be of the form 3^x * 37^y. For their product to be 999, the sum of exponents for each prime must match:

• x1 + x2 + x3 + x4 = 3 (for prime 3, x_i ≥ 0)

• y1 + y2 + y3 + y4 = 1 (for prime 37, y_i ≥ 0)

Using the Stars and Bars formula (n + k - 1) C (k - 1):

• For x (n=3 items, k=4 bins): (3 + 4 - 1) C (4 - 1) = 6 C 3 = 20 ways.

• For y (n=1 item, k=4 bins): (1 + 4 - 1) C (4 - 1) = 4 C 3 = 4 ways.
  The total number of such tuples is the product of these independent ways: 20 * 4 = 80.

Problem 4: ISI UGA Counting Problem

In an n-question test, at least 'i' questions were wrongly answered by 2^(n-i) students. If the total number of wrong answers is 2047, find n.

Let N_i be the number of students who wrongly answered at least i questions (N_i = 2^(n-i)).

To find the total wrong answers, we first convert "at least" to "exactly": E_k = N_k - N_(k+1) = 2^(n-k) - 2^(n-(k+1)). For exactly 'n' wrong answers, E_n = N_n = 2^(n-n) = 1 student.

The total wrong answers (T) = Σ (k * E_k) for k = 1 to n. Expanding and collecting terms, this sum simplifies to a geometric progression:

T = (1 * E_1) + (2 * E_2) + … + (n * E_n)

T = 2^(n-1) + 2^(n-2) + … + 2^1 + 2^0 = 2^n - 1.

Given T = 2047, we have 2^n - 1 = 2047 => 2^n = 2048.

Since 2^11 = 2048, n = 11.

(Memory Tip: Convert "at least" counts to "exactly" counts, then sum to simplify into a geometric series.)

Problem 5: ISI UGB 2025 Perfect Cube Problem

Given nine distinct ordered pairs (a_i, b_i) in N x N, forming set S = {2^a_i * 3^b_i | i = 1, …, 9}. Prove there exist three distinct elements in S whose product is a perfect cube.

For the product of three elements from S to be a perfect cube, the sum of their exponents for each prime (2 and 3) must be a multiple of 3. That is:

• (a_i1 + a_i2 + a_i3) ≡ 0 (mod 3)

• (b_i1 + b_i2 + b_i3) ≡ 0 (mod 3)

Each element (2^a_i * 3^b_i) can be classified by its exponent remainders modulo 3: (a_i mod 3, b_i mod 3). There are 3 * 3 = 9 possible distinct patterns: (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2).

Since we have nine distinct elements and nine distinct patterns, each element must correspond to a unique pattern. By the Pigeonhole Principle, we can always select three elements whose (mod 3) exponent sums for both a and b are (0,0). 

For instance, consider the elements corresponding to patterns (0,0), (1,1), and (2,2). Their a exponents sum to 0+1+2 = 3 ≡ 0 (mod 3), and similarly for their b exponents. Thus, a triplet forming a perfect cube product must exist.

Problem 6: Flowless Permutations Problem

A permutation a1, ..., a5 of {1, 2, 3, 4, 5} is "flowless" if no a_i, a_j, a_k (i < j < k) form an arithmetic progression (AP). All 3-term APs in {1,2,3,4,5} (e.g., (1,2,3), (1,3,5)) include the number 3. Thus, 3's position is critical in determining flowless permutations.

Analysis by Position of 3:

• Case 1: 3 at a1 (First Position): To avoid APs, the remaining numbers {1,2,4,5} must be arranged as (5,4) and (1,2) relative to each other. Result: 6 permutations.

• Case 2: 3 at a2 (Second Position): 0 flowless permutations. Any arrangement will form an AP.

• Case 3: 3 at a3 (Third Position - Middle): To avoid APs, pairs (1,5) and (2,4) must be on opposite sides of 3. This leads to two sub-cases: (1,5) left, (2,4) right (2! * 2! = 4 ways); or (2,4) left, (1,5) right (2! * 2! = 4 ways). Total: 8 permutations.

• Case 4: 3 at a4 (Fourth Position): 0 flowless permutations (symmetrical to Case 2).

• Case 5: 3 at a5 (Fifth Position - Last): Symmetrical to Case 1. The remaining {1,2,4,5} must be ordered as (2,1) and (4,5) relative to each other. Result: 6 permutations.

Total Flowless Permutations: Summing results from all cases: 6 + 0 + 8 + 0 + 6 = 20.

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