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Combinatorial Counting and Bijections PYQs for GATE DA

Solve 2+ Combinatorial Counting and Bijections previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

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Question 1

2026 PYQ

Level 3: Exam Standard

The number of bijections f (\cdot) from the set S = {1, 2, 3, 4} to itself such that f (f (n)) = n, for all n \in S, is __________ . (Answer in integer)

Question 2

2024 PYQ

Level 3: Exam Standard

How many 4-digit positive integers divisible by 3 can be formed using only the digits {1, 3,4, 6, 7}, such that no digit appears more than once in a number?

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Combinatorial Counting and Bijections PYQs for GATE DA

Solve 2+ Combinatorial Counting and Bijections previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Combinatorial Counting and Bijections

Chapter Roadmap

Combinatorial Counting and Bijections

Topic 1: Digit-Based Counting

Forming numbers from restricted sets using divisibility rules (mod 3/9).

Topic 2: Bijections & Involutions

Counting self-inverse mappings $f (f (n)) = n$ .

Goal: Master modular arithmetic for filtering combinations and permutation adjustments for leading zeros.

The Core Idea: Divisibility by 3

Number is divisible by 3

$⟺$

Sum of digits is divisible by 3

This rule allows us to ignore digit positions initially. We only care about which combination of digits sums to a multiple of 3.

Digits

{1, 3, 4, 7}

→

Sum

1 + 3 + 4 + 7 = 15

Combinatorial Counting and Bijections: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Aptitude · 2026 NAT

The number of bijections $f (\cdot)$ from the set $S = {1, 2, 3, 4}$ to itself such that $f (f (n)) = n$ , for all $n \in S$ , is __________ . (Answer in integer)

Correct Answer:

10.00

Step-by-Step Solution

Insight: The condition $f (f (n)) = n$ defines an involution, meaning the permutation consists entirely of 1-cycles (fixed points) and 2-cycles (swaps).

Exam route: Use the involution recurrence $I (n) = I (n - 1) + (n - 1) I (n - 2)$ . With $I (1) = 1$ and $I (2) = 2$ , we get $I (3) = 2 + 2 (1) = 4$ , and $I (4) = 4 + 3 (2) = 10$ .

Learning route:

We classify the bijections by their cycle structure for $n = 4$ :

Zero swaps (4 fixed points): There is exactly $(0 4) = 1$ way.
One swap (2 fixed points): Choose 2 elements to swap out of 4. This is $(2 4) = 6$ ways.
Two swaps (0 fixed points): Choose 2 elements for the first swap ( $(2 4) = 6$ ), and the remaining 2 form the second swap ( $(2 2) = 1$ ). Since the two swaps are indistinguishable, we divide by $2!$ . This gives $\frac{6 \times 1}{2} = 3$ ways.

Total involutions = $1 + 6 + 3 = 10$ .

Common Trap: Forgetting to divide by $2!$ in the two-swap case leads to $6 \times 1 = 6$ ways, incorrectly totaling $1 + 6 + 6 = 13$ .

Verification: The recurrence $I (4) = I (3) + 3 I (2) = 4 + 3 (2) = 10$ perfectly matches the manual enumeration.

Question 2 · Quantitative Aptitude · 2024 MCQ

How many 4-digit positive integers divisible by 3 can be formed using only the
digits {1, 3,4, 6, 7}, such that no digit appears more than once in a number?

A.
24
B.
48
C.
72
D.
12

Correct Answer:

Step-by-Step Solution

Insight: A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of all 5 available digits is 21. To form a 4-digit number, we must drop exactly one digit, and that dropped digit must be a multiple of 3.

Exam route: The multiples of 3 in ${1, 3, 4, 6, 7}$ are 3 and 6.

Case 1: Drop 3. Digits are ${1, 4, 6, 7}$ . All are non-zero, so $4! = 24$ ways.

Case 2: Drop 6. Digits are ${1, 3, 4, 7}$ . All are non-zero, so $4! = 24$ ways.

Total = $24 + 24 = 48$ .

Learning route:

Check the sum of the given set: $1 + 3 + 4 + 6 + 7 = 21$ .
Since 21 is divisible by 3, removing a digit $x$ leaves a sum of $21 - x$ .
For the remaining 4 digits to be divisible by 3, $21 - x$ must be a multiple of 3, which means $x$ itself must be a multiple of 3.
The available multiples of 3 in the set are 3 and 6. Thus, we have exactly two valid subsets of 4 digits: ${1, 4, 6, 7}$ and ${1, 3, 4, 7}$ .
Neither subset contains the digit 0, so there are no leading-zero constraints to worry about. Each subset can form $4! = 24$ valid 4-digit numbers.
Total valid numbers = $24 + 24 = 48$ .

Common Trap: Forgetting that 0 is not in the set and unnecessarily subtracting leading zero cases, or failing to realize that dropping a non-multiple of 3 ruins the divisibility.

Verification: Both subsets sum to a multiple of 3 (18 and 15), and both yield 24 permutations. $24 + 24 = 48$ .

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Combinatorial Counting and Bijections PYQs for GATE DA

Chapter Roadmap: Combinatorial Counting and Bijections

Chapter Roadmap

Topic 1: Digit-Based Counting

Topic 2: Bijections & Involutions

The Core Idea: Divisibility by 3

The Core Idea: Divisibility by 3

Combinatorial Counting and Bijections: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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