Counting, Sets and Elementary Probability PYQs for GATE CS

Solve 3+ Counting, Sets and Elementary Probability previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Counting, Sets and Probability

Your Learning Journey

1 Sets and Inclusion-Exclusion

Organize objects into collections and count them systematically. Learn the fundamental principle for avoiding double-counting.

Weightage: ~40% of chapter questions

2 Permutations and Arrangements

Count ordered selections. Master factorial notation and arrangement problems.

Weightage: ~40% of chapter questions

3 Elementary Dice Probability

Apply counting to chance events. Calculate probabilities for dice rolls and similar experiments.

Weightage: ~20% of chapter questions

What You Will Master

Systematic organization of counting problems
Core formulas for sets, permutations, and probability
Ability to recognize question patterns quickly
Techniques to avoid common calculation errors

What Are Sets? The Foundation of Counting

Definition

A set is a well-defined collection of distinct objects.

Notation

Sets: $A, B, C, \dots$
Elements: $a, b, c, \dots$
" $x$ belongs to $A$ ": $x \in A$
" $x$ does not belong to $A$ ": $x \in / A$

Examples

$A = {2, 4, 6, 8}$ (even numbers less than 10)
$B = {a, e, i, o, u}$ (vowels in English)
$C = {1, 2, 3, \dots, 100}$ (first 100 natural numbers)

Cardinality

The number of elements in a set $A$ is denoted by:
$n (A) or ∣ A ∣$

For the examples above:

$n (A) = 4$
$n (B) = 5$
$n (C) = 100$

Why Sets Matter in Counting

Sets provide a systematic way to:

Group objects by shared properties
Avoid counting the same object twice
Apply formulas to find totals efficiently

Counting, Sets and Elementary Probability: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Quantitative Aptitude · 2026_Set1 MCQ

An unbiased six-faced dice whose faces are marked with numbers 1, 2, 3, 4, 5, and 6 is rolled twice in succession and the number on the top face is recorded each time. The probability that the number appearing in the second roll is an integer multiple of the number appearing in the first roll is __________

A.
$\frac{1}{6}$
B.
$\frac{5}{18}$
C.
$\frac{7}{18}$
D.
$\frac{5}{6}$

Correct Answer:

Step-by-Step Solution

Insight: fix the first roll, count the multiples available to the second roll; "integer multiple" includes equality.

Exam route:

Sample space $∣ S ∣ = 6 \times 6 = 36$ ordered pairs $(a, b)$ .
For each first roll $a$ , count $b \in {1, \dots, 6}$ with $b = k a$ :
$a = 1$ : $b \in {1, 2, 3, 4, 5, 6}$ → 6
$a = 2$ : $b \in {2, 4, 6}$ → 3
$a = 3$ : $b \in {3, 6}$ → 2
$a = 4$ : $b \in {4}$ → 1
$a = 5$ : $b \in {5}$ → 1
$a = 6$ : $b \in {6}$ → 1
Favourable $= 6 + 3 + 2 + 1 + 1 + 1 = 14$ .
$P = 14/36 = 7/18$ .

Learning route:

Let the first roll be $a$ and the second be $b$ . The condition " $b$ is an integer multiple of $a$ " means $b = k a$ for some positive integer $k$ . Since $1 \leq b \leq 6$ , the multiples of $a$ not exceeding $6$ are exactly $a, 2 a, 3 a, \dots$ up to $6$ .

Build the case table:

| $a$ | Allowed $b$ | Count |

|---|---|---|

| 1 | 1, 2, 3, 4, 5, 6 | 6 |

| 2 | 2, 4, 6 | 3 |

| 3 | 3, 6 | 2 |

| 4 | 4 | 1 |

| 5 | 5 | 1 |

| 6 | 6 | 1 |

Total favourable $= 6 + 3 + 2 + 1 + 1 + 1 = 14$ . Since the two rolls are independent and ordered, $∣ S ∣ = 36$ , so

$P = \frac{14}{36} = \frac{7}{18} .$

Verification: the complement (second roll is NOT a multiple of the first) has $36 - 14 = 22$ outcomes; $22/36 = 11/18$ , and $7/18 + 11/18 = 1$ .

Wrong-path autopsy:

$1/6$ (A) counts only the $a = 1$ row (6 outcomes) and forgets the other five rows.
$5/18$ (B) drops equality — counting only <i>strict</i> multiples — giving $5 + 2 + 1 + 0 + 0 + 0 = 8$ or a similar miscount; the problem says "integer multiple", which includes $b = a$ .
$5/6$ (D) is the complement of $1/6$ , i.e. the same miscount promoted to the other side.

Generalisation: for any condition linking the first and second rolls, fix one roll and enumerate the compatible values of the other; never treat the 11 possible sums or the 21 unordered pairs as equally likely.

Question 2 · Quantitative Aptitude · 2024_Set2 MCQ

Two wizards try to create a spell using all the four elements, water, air, fire, and earth. For this, they decide to mix all these elements in all possible orders. They also decide to work independently. After trying all possible combination of elements, they conclude that the spell does not work.

How many attempts does each wizard make before coming to this conclusion, independently?

A.
24
B.
48
C.
16
D.
12

Correct Answer:

Step-by-Step Solution

Insight: "all possible orders of all four elements" is the textbook trigger for $4!$ ; "each wizard independently" does not multiply the per-wizard count.

Exam route:

Four distinct elements, all used, order matters $⟹ 4! = 24$ arrangements.
Each wizard works independently and tries all of them, so each wizard makes $24$ attempts.

Learning route:

The problem asks for the number of attempts <i>each</i> wizard makes. The two wizards are independent actors; the question is not asking for the total across both wizards.

Step 1 — identify the counting task. We are arranging all four distinct elements (water, air, fire, earth) in a sequence. Order matters because "water, air, fire, earth" is a different mix from "air, water, fire, earth".

Step 2 — apply the permutation-of-all formula. Arranging $n$ distinct objects in order gives $n!$ outcomes. Here $n = 4$ , so

$4! = 4 \times 3 \times 2 \times 1 = 24.$

Step 3 — read the question carefully. "How many attempts does each wizard make, independently?" Each wizard runs through all $24$ arrangements on their own. The presence of a second wizard does not change the count per wizard.

Verification: listing a few arrangements confirms the scale — starting with water there are $3! = 6$ arrangements, and there are $4$ choices for the first element, giving $4 \times 6 = 24$ .

Wrong-path autopsy:

$48$ (B) multiplies $24 \times 2$ , confusing "each wizard" with "both wizards together".
$16$ (C) uses $4^{2}$ or $2^{4}$ , mixing up permutations with independent binary choices.
$12$ (D) uses $4! /2$ or $4 \times 3$ , either halving for a non-existent symmetry or stopping one step early.

Generalisation: whenever a problem says "arrange all $n$ distinct objects in all possible orders", the answer is $n!$ ; extra actors working independently do not change the per-actor count unless the question explicitly asks for a total.

Question 3 · Quantitative Aptitude · 2024_Set2 MCQ

In an engineering college of 10,000 students, 1,500 like neither their core branches nor other branches. The number of students who like their core branches is

1/ 4^{th}

of the number of students who like other branches. The number of students who like both their core and other branches is 500.

The number of students who like their core branches is

A.
1,800
B.
3,500
C.
1,600
D.
1,500

Correct Answer:

Step-by-Step Solution

Insight: "Neither" tells you the union immediately; the ratio then collapses to one unknown.

Exam route:

Total $= 10, 000$ , neither $= 1, 500 ⟹ n (C \cup O) = 10, 000 - 1, 500 = 8, 500$ .
Given $n (C) = \frac{1}{4} n (O)$ , so $n (O) = 4 n (C)$ .
Inclusion-exclusion: $n (C \cup O) = n (C) + n (O) - n (C \cap O)$ .
$8, 500 = n (C) + 4 n (C) - 500 = 5 n (C) - 500$ .
$5 n (C) = 9, 000 ⟹ n (C) = 1, 800$ .

Learning route:

Let $C$ be the set of students who like their core branch and $O$ the set who like other branches. The universal set has $n (U) = 10, 000$ .

Step 1 — use "neither" to get the union. Students who like neither are outside $C \cup O$ , so

$n (C \cup O) = n (U) - 1, 500 = 8, 500.$

Step 2 — translate the ratio. $n (C) = \frac{1}{4} n (O)$ means $n (O) = 4 n (C)$ .

Step 3 — apply inclusion-exclusion for two sets:

$n (C \cup O) = n (C) + n (O) - n (C \cap O) .$

Substituting $n (O) = 4 n (C)$ and $n (C \cap O) = 500$ :

$8, 500 = n (C) + 4 n (C) - 500 = 5 n (C) - 500.$

Step 4 — solve: $5 n (C) = 9, 000$ , so $n (C) = 1, 800$ .

Verification: $n (O) = 7, 200$ . Then $n (C) + n (O) - n (C \cap O) = 1, 800 + 7, 200 - 500 = 8, 500$ , and $8, 500 + 1, 500 = 10, 000$ . All constraints satisfied.

Wrong-path autopsy:

Choosing $3, 500$ (B) comes from computing $8, 500 - 5, 000$ or mixing up which set is $1/4$ of which; it breaks the ratio $n (C) = \frac{1}{4} n (O)$ .
Choosing $1, 600$ (C) comes from using $8, 000$ instead of $8, 500$ (forgetting to add the $500$ intersection back); this violates the union equation.
Choosing $1, 500$ (D) is just echoing the "neither" count — a comprehension slip, not a calculation.

Generalisation: whenever a problem gives a total, a "neither" count, and a ratio between two sets, convert "neither" to the union first, then substitute the ratio into inclusion-exclusion.

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Counting, Sets and Elementary Probability PYQs for GATE CS

Chapter Roadmap: Counting, Sets and Probability

What Are Sets? The Foundation of Counting

Definition

Notation

Examples

Cardinality

Why Sets Matter in Counting

Counting, Sets and Elementary Probability: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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