Permutation and Combination is an important topic in the Quantitative Aptitude section of the RBI Grade B Phase 1 exam. These questions test candidates’ ability to count possible arrangements or selections in different situations. They are based on basic counting principles, logic, and sometimes probability. The past exam analysis tells, Permutation and Combination questions usually carry 1 to 2 marks in the Quants section, which carries a total of 30 marks. The difficulty level of this part is moderate for most candidates. This means they can be solved in under 50 seconds if you know the right method or solving techniques. That gives you extra time for lengthy questions like Data Interpretation or complex arithmetic problems. In this article, we’ll discuss the basics, formulas, patterns, and easy tricks to solve Permutation and Combination questions quickly and accurately.
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The main difference between Permutation and Combination is that in Permutation, repetition or the order of things matter, but in Combination, they don’t. So, it’s important to know the types of them.
If you check RBI Grade B past papers, these are the main types you will see:
Here, you are asked to arrange objects in a certain order. Example: How many ways can the letters of the word “BANK” be arranged? These are the most basic and are quick to solve with factorials.
Some objects repeat. This reduces the number of arrangements.
Example: How many different words can be formed from “LETTER”?
Here, you divide by the factorial of each repeated letter count.
Here, certain conditions are given.
Example: In how many ways can 5 people stand so that two specific people always stand together?
These require grouping and then arranging.
Here, you select items without considering the order.
Example: From 10 players, choose 3 for a team.
Formula:
Some selections are not allowed or must follow a rule.
Example: From 8 candidates, choose 4, but two friends cannot be together in the selection.
These need careful case-wise counting.
These combine arrangements and selections.
For example: Select 3 students from the given 5, and then arrange them in a row.
These need both formulas below in one problem:
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There are a few formulas you must remember:
n!=n×(n−1)×(n−2)…3×2×1n! = n \times (n-1) \times (n-2) \dots 3 \times 2 \times 1
For example: 4!=244! = 24
Number of arrangements of rr items from nn items:
Meaning: total arrangements when some items repeat.
Number of selections of rr items from nn items:
When some items look the same:
If one task can be done in m ways and another in n ways, total ways = m × n.
Here’s the approach that works in the exam:
1. Write the Given Data Clearly: List out n and r first. Identify if it’s arrangement or selection.
2. Decide Between P and C Early: If the order matters, use permutation. If not, use combination.
3. Break Down Restrictions: If conditions are given, think in cases before applying formulas.
4. Use the Principle of Fundamental Counting: For step-by-step tasks, multiply the ways for each step.
5. Avoid Long Factorial Calculations: Cancel terms before multiplying to save time.
Given below are the types of Permutation and Combination
How many different 4-digit numbers can be formed from the digits 1, 2, 3, 4 without repetition?
Number of ways =
Answer: 24 numbers.
In how many ways can the letters of “BANK” be arranged so that B and A are always together?
Treat BA as one unit → units = BA, N, K = 3 units.
Arrangements = 3!=63! = 6
BA can be AB as well, so multiply by 2 → Total = 12 ways.
Answer: 12 arrangements.
From 8 students, choose 3 for a project.
Number of ways =
Answer: 56 ways.
From 8 candidates, choose 3, but two specific candidates cannot be together.
Total ways without restriction = 8C3=56^8C_3 = 56
Ways with both together: Treat them as 1 unit → select 1 more from 6 → 6C1=6^6C_1 = 6 ways.
Within the pair, they can be arranged in 1 way (since order doesn’t matter in selection).
So restricted ways = 6.
Allowed ways = 56 − 6 = 50 ways.
Answer: 50 ways.
Given below are the most important shortcuts that you must know to quickly solve all types of Permutation and Combination:
Daily practice is the only way to make this topic quick in the exam.
Try solving these questions below quickly:
Now, it’s time to practice the types of questions that you can face in the real exam. Go through all the questions below and try to solve them on your own. Compare your answers with the ANSWER KEYS and SOLUTIONS given under these questions.
Question 1: From a group of 8 men and 6 women, a committee of 8 persons is to be selected. Find the number of ways in which the committee can be formed with at-least 5 men in the group.
A) 1572
B) 1577
C) 1569
D) 1589
E) None of these
Question 2: Find total number of ways in which the word “UNCHATACTERISTICALLY” can be arranged so all the vowels always come together.
A) 13190 × 11!
B) 12740 × 11!
C) 12180 × 11!
D) 17910 × 11!
E) None of these
Question 3: The sum of all the possible numbers of four digits formed by digits 2, 3, 4 and 5 using each digit once is:
A) 94424
B) 99657
C) 93324
D) 94367
E) None of these
Question 4: How many four letter words having at most three consonants and at least 1 vowel can be formed by using the letters of the word “CASTLIEY”?
A) 65
B) 360
C) 720
D) 1560
E) None of these
Question 5: How many words with or without meaning can be formed by using all the letters of the words “PRACTICE” such that all the vowels always occupy even positions?
A) 240
B) 720
C) 1440
D) 2880
E) None of these
प्रश्न 1: 8 पुरुष और 6 महिला के एक समूह से, 8 व्यक्तियों के एक समिति का चयन किया जाता है|कम से कम 5 पुरुष के साथ समिति का चयन कितने तरीके से किया जाता है?
A) 1572
B) 1577
C) 1569
D) 1589
E) None of these
प्रश्न 2: उन तरीकों की कुल संख्या ज्ञात करें जिनमें शब्द “UNCHATACTERISTICALLY” को व्यवस्थित किया जा सकता है ताकि सभी vowels हमेशा एक साथ आ सकें।
A) 13190 × 11!
B) 12740 × 11!
C) 12180 × 11!
D) 17910 × 11!
E) None of these
प्रश्न 3: प्रत्येक अंक का एक बार प्रयोग करके अंक 2, 3, 4 और 5 से बनने वाली चार अंकों की सभी संभावित संख्याओं का योग है:
A) 94424
B) 99657
C) 93324
D) 94367
E) None of these
प्रश्न 4: शब्द “CASTLIEY” के अक्षरों का उपयोग कर ऐसे कितने 4 अक्षरों के शब्द का निर्माण किया जा सकता है जिसमे अधिकतम तीन consonants और कम से कम 1 vowel हो?
A) 65
B) 360
C) 720
D) 1560
E) None of these
ANSWER KEYS and SOLUTIONS:
| 1) – 4) | 2) – 2) | 3) – 3) | 4) – 4) | 5) – 3) |
Solution 1: 4)
Desired number of ways = (8C5 × 6C3) + (8C6 × 6C2) + (8C7 × 6C1) + 8C8 = 1120 + 420 + 48 + 1 = 1589
Hence, option d.
Solution 2: 2)
Desired number of ways {7!/(3! × 2!)} × {14!/(3! × 3! × 2!)} = 12740 × 11!
Hence, option b.
Solution 3: 3)
Total number of different numbers of 4 digits formed using these digits only once = 4! = 24
The number formed whose unit digit is ‘2’ = 6
The number formed whose unit digit is ‘3’ = 6
The number formed whose unit digit is ‘4’ = 6
The number formed whose unit digit is ‘5’ = 6
Sum of the total number of unit digits = 2 × 6 + 3 × 6 + 4 × 6 + 5 × 6 = 84
Desired sum = 84 + 84 × 10 + 84 × 100 + 84 × 1000 = 84 × (1000 + 100 + 10 + 1) = 84 × 1111 = 93324
Hence, option c.
Solution 4: 4)
There are 5 consonants and 3 vowels in the word “CASTLIEY”
Case I: 1 vowel and 3 consonants are there in the word
Number of ways of selecting letters = 3C1 × 5C3 = 3 × 10 = 30
Case II: 2 vowels and 2 consonants are there in the word
Number of ways of selecting letters = 3C2 × 5C2 = 3 × 10 = 30
Case III: 3 vowels and 1 consonant are there in the word
Number of ways of selecting letters = 3C3 × 5C1 = 1 × 5 = 5
So the total number of ways in which the letter are selected = 30 + 30 + 5 = 65
These 4 letters can be arranged with in the word in 4! = 24 ways
So the total number of words formed = 24 × 65 = 1560
Hence, option d.
Solution 5: 3)
There are 4 even positions and 3 vowels.
Number of ways of filling these 3 vowels = 4P3 = 24
Number of ways of filling 5 consonants = 5!/2! = 60
So, the total number of words formed = 60 × 24 = 1440
Hence, option c.
ALSO READ: Word Swap Tricks for RBI Grade B Phase 1 Exam
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Permutation and Combination is a scoring topic in the Quant section of the RBI Grade B exam if you master the basic formulas and shortcuts. Once you get comfortable, you can solve these questions in seconds. Make them a fixed part of your daily Quant practice, and you will secure these marks quickly in the real exam.
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