Counting, Sets and Elementary Probability Practice Questions for GATE CS

Solve 48+ Counting, Sets and Elementary Probability practice 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 (10 Problems)

Question 1 · Quantitative Aptitude MCQ

In a group of 50 people, 30 like tea and 25 like coffee. What is the minimum number of people who like both tea and coffee?

A.
5
B.
15
C.
25
D.
55

Correct Answer:

Step-by-Step Solution

Key idea: Use the inclusion-exclusion principle and the fact that the union cannot exceed the total population.

Step 1: Let $T$ be the set of tea lovers and $C$ be the set of coffee lovers. We have $n (T) = 30$ , $n (C) = 25$ , and the total population $n (U) = 50$ .

Step 2: The inclusion-exclusion formula is $n (T \cup C) = n (T) + n (C) - n (T \cap C)$ .

Step 3: Since $n (T \cup C) \leq n (U)$ , we have $30 + 25 - n (T \cap C) \leq 50$ .

Step 4: Simplify the inequality: $55 - n (T \cap C) \leq 50 ⟹ n (T \cap C) \geq 5$ .

Answer: 5

Question 2 · Quantitative Aptitude MCQ

A fair die is rolled twice. Let $a$ be the first roll and $b$ be the second. Which of the following is the exact number of outcomes where $b \geq a$ ?

A.
15
B.
18
C.
21
D.
36

Correct Answer:

Step-by-Step Solution

Key idea: Fix the first roll $a$ and count the valid second rolls $b$ that satisfy $b \geq a$ .

Step 1: If $a = 1$ , $b$ can be 1, 2, 3, 4, 5, 6 (6 outcomes).

Step 2: If $a = 2$ , $b$ can be 2, 3, 4, 5, 6 (5 outcomes).

Step 3: If $a = 3$ , $b$ can be 3, 4, 5, 6 (4 outcomes).

Step 4: If $a = 4$ , $b$ can be 4, 5, 6 (3 outcomes).

Step 5: If $a = 5$ , $b$ can be 5, 6 (2 outcomes).

Step 6: If $a = 6$ , $b$ can be 6 (1 outcome).

Step 7: Sum the cases: $6 + 5 + 4 + 3 + 2 + 1 = 21$ .

Answer: 21

Question 3 · Quantitative Aptitude MCQ

Assertion (A): The number of outcomes where the second roll is an integer multiple of the first roll is 14.

Reason (R): For each first roll $a$ , the number of valid second rolls $b$ is the count of multiples of $a$ in ${1, 2, 3, 4, 5, 6}$ , which sums to $6 + 3 + 2 + 1 + 1 + 1 = 14$ .

A.
Both A and R are true and R is the correct explanation of A.
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: Verify the assertion by performing the case count described in the reason.

Step 1: Evaluate Assertion (A). We need to count pairs $(a, b)$ where $b$ is an integer multiple of $a$ .

Step 2: If $a = 1$ , $b \in {1, 2, 3, 4, 5, 6}$ (6 outcomes).

Step 3: If $a = 2$ , $b \in {2, 4, 6}$ (3 outcomes).

Step 4: If $a = 3$ , $b \in {3, 6}$ (2 outcomes).

Step 5: If $a = 4, 5, 6$ , $b$ can only be $a$ itself (1 outcome each).

Step 6: Sum the cases: $6 + 3 + 2 + 1 + 1 + 1 = 14$ . Thus, A is true.

Step 7: Evaluate Reason (R). The reason correctly describes the exact case-counting method and sum that proves A. Thus, R is true and is the correct explanation.

Answer: Both A and R are true and R is the correct explanation of A.

Question 4 · Quantitative Aptitude MCQ

Let $U$ be the set of integers from 1 to 30. Let $A$ be the set of all multiples of 4 in $U$ . What is the maximum possible number of elements in $A \cap B$ , where $B$ is a subset of $U$ with exactly 5 elements?

A.
5
B.
7
C.
12
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The intersection $A \cap B$ cannot be larger than the smaller of the two sets $A$ and $B$ .

Step 1: Find the elements of $A$ . Multiples of 4 from 1 to 30 are ${4, 8, 12, 16, 20, 24, 28}$ .

Step 2: Count the elements in $A$ . There are 7 elements, so $n (A) = 7$ .

Step 3: We are given that $B$ has exactly 5 elements, so $n (B) = 5$ .

Step 4: The maximum size of $A \cap B$ occurs when all elements of $B$ are also in $A$ . Since $n (B) = 5$ and $n (A) = 7 \geq 5$ , the maximum intersection is 5.

Answer: 5

Question 5 · Quantitative Aptitude MCQ

Assertion (A): The number of distinct arrangements of the letters in the word "BOOK" is 24.

Reason (R): The formula $^{4} P_{4} = 4!$ gives the number of ways to arrange 4 distinct objects in a row.

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is false but R is true
D.
A is false but R is false

Correct Answer:

Step-by-Step Solution

Key idea: The permutation formula $^{n} P_{n} = n!$ applies only to distinct objects.

Step 1: Evaluate Assertion (A). The word "BOOK" has 4 letters, but the letter 'O' repeats twice.

Step 2: The number of distinct arrangements is $\frac{4 !}{2 !} = \frac{24}{2} = 12$ . Thus, A is false.

Step 3: Evaluate Reason (R). The statement " $^{4} P_{4} = 4!$ gives the number of ways to arrange 4 distinct objects" is a mathematically true statement.

Step 4: Since A is false and R is true, the correct option is C.

Answer: A is false but R is true

Question 6 · Quantitative Aptitude MCQ

A fair six-faced die is rolled twice. What is the maximum possible number of ordered outcomes where the first roll is strictly greater than the second roll?

A.
12
B.
15
C.
18
D.
21

Correct Answer:

Step-by-Step Solution

Key idea: Use symmetry and the boundary condition where the two rolls are equal.

Step 1: The total number of ordered outcomes for two rolls is $6 \times 6 = 36$ .

Step 2: The outcomes can be split into three categories: first > second, first < second, and first = second.

Step 3: The number of outcomes where first = second is exactly 6: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

Step 4: The remaining outcomes are $36 - 6 = 30$ . By symmetry, exactly half of these have first > second.

Step 5: Calculate the result: $30/2 = 15$ .

Answer: 15

Question 7 · Quantitative Aptitude MCQ

Let $S$ be a set containing 20 elements. If a subset $A$ contains 8 elements, what is the cardinality of the complement of $A$ with respect to $S$ ?

A.
12
B.
28
C.
160
D.
-12

Correct Answer:

Step-by-Step Solution

Key idea: The complement of a subset $A$ with respect to a universal set $S$ contains all elements in $S$ that are not in $A$ .

Step 1: Identify the total number of elements in the universal set $S$ , which is $n (S) = 20$ .

Step 2: Identify the number of elements in the subset $A$ , which is $n (A) = 8$ .

Step 3: Apply the complement formula: $n (A^{'}) = n (S) - n (A)$ .

Step 4: Calculate the result: $20 - 8 = 12$ .

Answer: 12

Question 8 · Quantitative Aptitude MCQ

For any non-negative integers $n$ and $r$ such that $r \leq n$ , which of the following inequalities is always true?

A.
$^{n} P_{r} \geq n!$
B.
$^{n} P_{r} \leq n!$
C.
$^{n} P_{r} \geq (n + 1)!$
D.
$^{n} P_{r} = (n - r)!$

Correct Answer:

Step-by-Step Solution

Key idea: The permutation formula $^{n} P_{r}$ is a product of $r$ decreasing terms starting from $n$ .

Step 1: Recall the formula: $^{n} P_{r} = n \times (n - 1) \times \dots \times (n - r + 1)$ .

Step 2: Notice that $n! = n \times (n - 1) \times \dots \times 1$ .

Step 3: Since $r \leq n$ , the product for $^{n} P_{r}$ contains $r$ terms, while $n!$ contains $n$ terms.

Step 4: If $r = n$ , $^{n} P_{r} = n!$ . If $r < n$ , $^{n} P_{r}$ is missing the terms from $(n - r)$ down to 1, which are all $\geq 1$ . Thus, $^{n} P_{r} \leq n!$ .

Answer: $^{n} P_{r} \leq n!$

Question 9 · Quantitative Aptitude MCQ

In how many ways can 3 distinct books be arranged on a shelf from a collection of 5 distinct books?

A.
10
B.
120
C.
60
D.
125

Correct Answer:

Step-by-Step Solution

Key idea: This is a permutation problem because the order of the books on the shelf matters.

Step 1: Identify the total number of objects $n = 5$ and the number of objects to arrange $r = 3$ .

Step 2: Apply the permutation formula $^{n} P_{r} = \frac{n !}{( n - r )!}$ .

Step 3: Substitute the values: $^{5} P_{3} = \frac{5 !}{( 5 - 3 )!} = \frac{5 !}{2 !}$ .

Step 4: Calculate the result: $\frac{120}{2} = 60$ .

Answer: 60

Question 10 · Quantitative Aptitude MCQ

A fair die is rolled twice. What is the minimum number of outcomes that must be removed from the sample space of 36 to ensure that no outcome contains a 6?

A.
10
B.
12
C.
25
D.
11

Correct Answer:

Step-by-Step Solution

Key idea: Use the complement. Find the number of outcomes that do not contain a 6, and subtract from the total.

Step 1: The total number of outcomes is 36.

Step 2: An outcome contains no 6 if both rolls are from the set ${1, 2, 3, 4, 5}$ .

Step 3: The number of such outcomes is $5 \times 5 = 25$ .

Step 4: The number of outcomes to remove is the total minus the outcomes with no 6: $36 - 25 = 11$ .

Answer: 11

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Counting, Sets and Elementary Probability Practice Questions 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 (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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