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Aptitude Problems on Permutation and Combination - Tips and Tricks to Solve in IBPS PO and Clerk Exams with examples.

Aptitude Problems on Permutation and Combination – Tips and Tricks to Solve in IBPS PO and Clerk Exams with examples.

Hello Students,

I am Rahul Sir, and today we will discuss one of the most important topics in the Quantitative Aptitude section of IBPS PO and Clerk examinations – Permutation and Combination. Many banking aspirants find this topic difficult because it involves counting arrangements and selections without actually listing all possibilities. However, once you understand the basic concepts and formulas, these questions become among the easiest scoring areas in competitive exams.

Permutation and Combination questions test your logical thinking, counting ability, and understanding of arrangements. They are frequently asked in IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, RRB PO, and other banking examinations. Usually, 1-3 questions appear directly or indirectly from this topic.

In this article, I will explain important concepts, formulas, shortcuts, and exam-oriented tricks with examples. By mastering these techniques, you can solve questions quickly and accurately during the exam.


1. Understanding Permutation and Combination Basics

Before learning shortcuts, it is important to understand the difference between permutation and combination.

Permutation

Permutation refers to arrangements where order matters.

Formula:

nPr = n! / (n-r)!

Combination

Combination refers to selections where order does not matter.

Formula:

nCr = n! / [r!(n-r)!]

Example

How many ways can 3 students be selected from 5 students?

Combination:

5C3 = 5! / (3! × 2!)

= 10

How many ways can President, Vice President, and Secretary be selected from 5 students?

Permutation:

5P3 = 5! / 2!

= 60

Exam Tip

Whenever positions are different, use permutation.

Whenever only selection is required, use combination.

Quick Memory Trick

  • Selection → Combination
  • Arrangement → Permutation

Most exam mistakes happen because students confuse these two concepts.


2. Mastering Factorials Quickly

Factorials are the foundation of permutation and combination.

Formula

n! = n × (n-1) × (n-2) …

Examples

5! = 120

6! = 720

7! = 5040

Important Values

NumberFactorial
11
22
36
424
5120
6720
75040

Shortcut

Instead of calculating complete factorials, cancel terms.

Example:

8! / 6!

= (8×7×6!)/6!

= 56

Example Question

Find 10P2.

10P2

= 10!/8!

= 10 × 9

= 90

Exam Strategy

Memorize factorial values up to 10!.

This saves significant time during calculations.


3. Circular Arrangement Questions

Questions involving people sitting around a table are common.

Formula

For n persons sitting in a circle:

Arrangement = (n-1)!

Example

How many ways can 6 persons sit around a round table?

=(6−1)!

=5!

=120

Why?

In circular arrangements, rotations are considered identical.

Example

A B C D and B C D A are same arrangements in a circle.

Quick Tip

SituationFormula
Straight Linen!
Circle(n-1)!

Example

8 persons around a table:

(8−1)!

=7!

=5040

This shortcut frequently appears in banking and insurance exams.


4. Arrangements with Repeated Letters

Many questions involve words containing repeated letters.

Formula

n! / (p! × q! × r!)

Where p, q, r are repeated letters.

Example

Find arrangements of “LEVEL”

Letters = 5

Repeated:

L = 2

E = 2

Answer:

5!/(2!×2!)

=120/4

=30

Example

Find arrangements of “BANANA”

Total letters = 6

A = 3

N = 2

Answer:

6!/(3!×2!)

=720/12

=60

Shortcut

Always identify repeated letters before applying factorial.

Ignoring repetitions leads to incorrect answers.


5. Selection of Committees

Committee questions are pure combination-based problems.

Example

A committee of 4 members is to be formed from 10 people.

10C4

=210

Example

Choose 2 boys from 5 and 3 girls from 7.

Required:

5C2 × 7C3

=10 × 35

=350

Shortcut

When separate groups exist:

Use multiplication.

Formula

Total Ways

= Ways from Group A × Ways from Group B

Committee questions are among the easiest permutation-combination questions in IBPS exams.


6. Arrangements with Restrictions

Many questions contain conditions.

Example

How many ways can 5 persons sit if A and B sit together?

Treat A and B as one unit.

Units become:

AB, C, D, E

Total units = 4

Arrangement:

4!

Internal arrangement of AB:

2!

Answer:

4! × 2!

=24 × 2

=48

Shortcut

When people stay together:

  1. Treat them as one object.
  2. Arrange units.
  3. Multiply internal arrangements.

This method solves most restriction-based questions quickly.


7. Arrangements Where Persons Cannot Sit Together

Opposite of the previous concept.

Example

Arrange A, B, C, D, E where A and B cannot sit together.

Method

Total arrangements

=5!

=120

Arrangements where A and B sit together

=4! × 2!

=48

Required

=120−48

=72

Shortcut Formula

Required

= Total Arrangements − Restricted Arrangements

This method saves time compared to direct counting.


8. Formation of Numbers Using Digits

Questions often involve forming numbers from digits.

Example

How many 3-digit numbers can be formed using digits 1,2,3,4,5?

Use permutation.

5P3

=5×4×3

=60

Example

How many 4-digit numbers can be formed using 0,1,2,3,4?

First digit cannot be zero.

Choose first digit:

4 ways

Remaining:

4P3

=24

Total

=4×24

=96

Shortcut

Whenever zero exists:

Handle first digit separately.

This avoids common mistakes.


9. Shortcut Techniques for IBPS Exams

Important Results

nC0 = 1

nCn = 1

nC1 = n

nCn−1 = n

Symmetry Property

nCr = nCn−r

Example:

10C8 = 10C2

=45

Benefit

Always choose the smaller value of r.

Example:

15C13

Instead solve

15C2

=105

Much faster.

Exam Trick

Reduce calculations by applying symmetry before solving.

This saves valuable exam time.


10. Previous Year IBPS Style Questions

Question 1

Select 3 members from 8 persons.

8C3

=56

Question 2

Arrange 5 books on a shelf.

5!

=120

Question 3

How many ways can 7 students stand in a row?

7!

=5040

Question 4

How many ways can 2 boys and 3 girls be selected from 5 boys and 6 girls?

5C2 × 6C3

=10 × 20

=200

Question 5

Arrange the letters of “BOOK”.

4! / 2!

=12

Exam Advice

Practice at least 50 mixed-level questions before the exam.

The more patterns you solve, the faster you identify the correct approach.


How Teachers from OdTutor Can Help

At OdTutor, experienced aptitude faculty help students master Permutation and Combination through concept-based learning, shortcut techniques, and extensive practice sessions. Our teachers explain difficult concepts using simple real-life examples and exam-oriented methods. Students receive topic-wise worksheets, doubt-solving sessions, mock tests, and previous year question discussions. Whether you are preparing for IBPS PO, IBPS Clerk, SBI, RRB, SSC, or other competitive exams, OdTutor provides personalized guidance to improve speed and accuracy. With regular practice and expert mentorship, students can confidently solve permutation and combination questions in the actual examination.


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