<\/p>\n\n\n\n
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. <\/strong>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\u2019ll discuss the basics, formulas, patterns, and easy tricks to solve Permutation and Combination questions quickly and accurately.<\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n ALSO READ: How to Score Maximum Marks in Cloze Test<\/a><\/strong><\/p>\n\n\n\n <\/p>\n\n\n\n The main difference between Permutation and Combination is that in Permutation, repetition or the order of things matter, but in Combination, they don\u2019t. So, it\u2019s important to know the types of them. <\/p>\n\n\n\n If you check RBI Grade B past papers, these are the main types you will see:<\/p>\n\n\n\n Here, you are asked to arrange objects in a certain order. Example: How many ways can the letters of the word \u201cBANK\u201d be arranged? These are the most basic and are quick to solve with factorials.<\/p>\n\n\n\n Some objects repeat. This reduces the number of arrangements.<\/p>\n\n\n\n Example: How many different words can be formed from \u201cLETTER\u201d?<\/p>\n\n\n\n Here, you divide by the factorial of each repeated letter count.<\/p>\n\n\n\n Here, certain conditions are given.<\/p>\n\n\n\n Example: In how many ways can 5 people stand so that two specific people always stand together?<\/strong><\/p>\n\n\n\n These require grouping and then arranging.<\/p>\n\n\n\n Here, you select items without considering the order.<\/p>\n\n\n\n Example: From 10 players, choose 3 for a team.<\/p>\n\n\n\n Formula:<\/strong><\/p>\n\n\n\n Some selections are not allowed or must follow a rule.<\/p>\n\n\n\n Example: From 8 candidates, choose 4, but two friends cannot be together in the selection.<\/p>\n\n\n\n These need careful case-wise counting.<\/p>\n\n\n\n These combine arrangements and selections.<\/p>\n\n\n\n For example: Select 3 students from the given 5, and then arrange them in a row.<\/strong><\/p>\n\n\n\n These need both formulas below in one problem<\/strong>:<\/p>\n\n\n\n <\/p>\n\n\n\n ALSO READ: How to Score Maximum Marks in Cloze Test<\/a><\/strong> & Ratio & Proportion Concepts and Shortcuts for RBI Grade B Exam<\/a><\/strong><\/p>\n\n\n\n <\/p>\n\n\n\n There are a few formulas you must remember:<\/p>\n\n\n\n n!=n\u00d7(n\u22121)\u00d7(n\u22122)\u20263\u00d72\u00d71n! = n \\times (n-1) \\times (n-2) \\dots 3 \\times 2 \\times 1<\/p>\n\n\n\n For example: 4!=244! = 24<\/p>\n\n\n\n Number of arrangements of rr items from nn items:<\/strong><\/p>\n\n\n\n Meaning: total arrangements when some items repeat.<\/p>\n\n\n\n Number of selections of rr items from nn items:<\/strong><\/p>\n\n\n\n When some items look the same:<\/p>\n\n\n\n If one task can be done in m ways and another in n ways, total ways = m \u00d7 n.<\/strong><\/p>\n\n\n\n Here\u2019s the approach that works in the exam:<\/p>\n\n\n\n 1. Write the Given Data Clearly: <\/strong>List out n and r first. Identify if it\u2019s arrangement or selection.<\/p>\n\n\n\n 2. Decide Between P and C Early: <\/strong>If the order matters, use permutation. If not, use combination.<\/p>\n\n\n\n 3. Break Down Restrictions:<\/strong> If conditions are given, think in cases before applying formulas.<\/p>\n\n\n\n 4. Use the Principle of Fundamental Counting: <\/strong>For step-by-step tasks, multiply the ways for each step.<\/p>\n\n\n\n 5. Avoid Long Factorial Calculations:<\/strong> Cancel terms before multiplying to save time.<\/p>\n\n\n\n Given below are the types of Permutation and Combination<\/p>\n\n\n\n How many different 4-digit numbers can be formed from the digits 1, 2, 3, 4 without repetition?<\/strong><\/p>\n\n\n\n Number of ways = <\/strong><\/p>\n\n\n\n Answer: 24 numbers.<\/strong><\/p>\n\n\n\n In how many ways can the letters of \u201cBANK\u201d be arranged so that B and A are always together?<\/p>\n\n\n\n Treat BA as one unit \u2192 units = BA, N, K = 3 units. Answer: 12 arrangements.<\/strong><\/p>\n\n\n\n From 8 students, choose 3 for a project.<\/strong><\/p>\n\n\n\n Number of ways = <\/strong><\/p>\n\n\n\n Answer: 56 ways.<\/strong><\/p>\n\n\n\n From 8 candidates, choose 3, but two specific candidates cannot be together.<\/p>\n\n\n\n Total ways without restriction = 8C3=56^8C_3 = 56<\/p>\n\n\n\n Ways with both together: Treat them as 1 unit \u2192 select 1 more from 6 \u2192 6C1=6^6C_1 = 6 ways.<\/p>\n\n\n\n Within the pair, they can be arranged in 1 way (since order doesn\u2019t matter in selection).<\/p>\n\n\n\n So restricted ways = 6.<\/p>\n\n\n\n Allowed ways = 56 \u2212 6 = 50 ways.<\/p>\n\n\n\n Answer: 50 ways.<\/strong><\/p>\n\n\n\n Given below are the most important shortcuts that you must know to quickly solve all types of Permutation and Combination:<\/p>\n\n\n\n Daily practice is the only way to make this topic quick in the exam.<\/p>\n\n\n\n Try solving these questions below quickly:<\/p>\n\n\n\n <\/p>\n\n\n\n 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<\/strong> given under these questions.<\/p>\n\n\n\n Question 1:<\/strong> 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.<\/p>\n\n\n\n A) 1572<\/p>\n\n\n\n B) 1577<\/p>\n\n\n\n C) 1569<\/p>\n\n\n\n D) 1589<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n Question 2:<\/strong> Find total number of ways in which the word \u201cUNCHATACTERISTICALLY\u201d can be arranged so all the vowels always come together.<\/p>\n\n\n\n A) 13190 \u00d7 11!<\/p>\n\n\n\n B) 12740 \u00d7 11!<\/p>\n\n\n\n C) 12180 \u00d7 11!<\/p>\n\n\n\n D) 17910 \u00d7 11!<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n Question 3:<\/strong> The sum of all the possible numbers of four digits formed by digits 2, 3, 4 and 5 using each digit once is:<\/p>\n\n\n\n A) 94424<\/p>\n\n\n\n B) 99657<\/p>\n\n\n\n C) 93324<\/p>\n\n\n\n D) 94367<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n Question 4:<\/strong> 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 \u201cCASTLIEY\u201d?<\/p>\n\n\n\n A) 65<\/p>\n\n\n\n B) 360<\/p>\n\n\n\n C) 720<\/p>\n\n\n\n D) 1560<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n Question 5:<\/strong> How many words with or without meaning can be formed by using all the letters of the words \u201cPRACTICE\u201d such that all the vowels always occupy even positions?<\/p>\n\n\n\n A) 240<\/p>\n\n\n\n B) 720<\/p>\n\n\n\n C) 1440<\/p>\n\n\n\n D) 2880<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n \u092a\u094d\u0930\u0936\u094d\u0928<\/strong> 1:<\/strong> 8 \u092a\u0941\u0930\u0941\u0937 \u0914\u0930 6 \u092e\u0939\u093f\u0932\u093e \u0915\u0947 \u090f\u0915 \u0938\u092e\u0942\u0939 \u0938\u0947, 8 \u0935\u094d\u092f\u0915\u094d\u0924\u093f\u092f\u094b\u0902 \u0915\u0947 \u090f\u0915 \u0938\u092e\u093f\u0924\u093f \u0915\u093e \u091a\u092f\u0928 \u0915\u093f\u092f\u093e \u091c\u093e\u0924\u093e \u0939\u0948|\u0915\u092e \u0938\u0947 \u0915\u092e 5 \u092a\u0941\u0930\u0941\u0937 \u0915\u0947 \u0938\u093e\u0925 \u0938\u092e\u093f\u0924\u093f \u0915\u093e \u091a\u092f\u0928 \u0915\u093f\u0924\u0928\u0947 \u0924\u0930\u0940\u0915\u0947 \u0938\u0947 \u0915\u093f\u092f\u093e \u091c\u093e\u0924\u093e \u0939\u0948?<\/p>\n\n\n\n A) 1572<\/p>\n\n\n\n B) 1577<\/p>\n\n\n\n C) 1569<\/p>\n\n\n\n D) 1589<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n \u092a\u094d\u0930\u0936\u094d\u0928<\/strong> 2:<\/strong> \u0909\u0928 \u0924\u0930\u0940\u0915\u094b\u0902 \u0915\u0940 \u0915\u0941\u0932 \u0938\u0902\u0916\u094d\u092f\u093e \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u091c\u093f\u0928\u092e\u0947\u0902 \u0936\u092c\u094d\u0926 “UNCHATACTERISTICALLY” \u0915\u094b \u0935\u094d\u092f\u0935\u0938\u094d\u0925\u093f\u0924 \u0915\u093f\u092f\u093e \u091c\u093e \u0938\u0915\u0924\u093e \u0939\u0948 \u0924\u093e\u0915\u093f \u0938\u092d\u0940 vowels \u0939\u092e\u0947\u0936\u093e \u090f\u0915 \u0938\u093e\u0925 \u0906 \u0938\u0915\u0947\u0902\u0964<\/p>\n\n\n\n A) 13190 \u00d7 11!<\/p>\n\n\n\n B) 12740 \u00d7 11!<\/p>\n\n\n\n C) 12180 \u00d7 11!<\/p>\n\n\n\n D) 17910 \u00d7 11!<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n \u092a\u094d\u0930\u0936\u094d\u0928<\/strong> 3:<\/strong> \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0905\u0902\u0915 \u0915\u093e \u090f\u0915 \u092c\u093e\u0930 \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0915\u0947 \u0905\u0902\u0915 2, 3, 4 \u0914\u0930 5 \u0938\u0947 \u092c\u0928\u0928\u0947 \u0935\u093e\u0932\u0940 \u091a\u093e\u0930 \u0905\u0902\u0915\u094b\u0902 \u0915\u0940 \u0938\u092d\u0940 \u0938\u0902\u092d\u093e\u0935\u093f\u0924 \u0938\u0902\u0916\u094d\u092f\u093e\u0913\u0902 \u0915\u093e \u092f\u094b\u0917 \u0939\u0948:<\/p>\n\n\n\n A) 94424<\/p>\n\n\n\n B) 99657<\/p>\n\n\n\n C) 93324<\/p>\n\n\n\n D) 94367<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n \u092a\u094d\u0930\u0936\u094d\u0928<\/strong> 4:<\/strong> \u0936\u092c\u094d\u0926 “CASTLIEY” \u0915\u0947 \u0905\u0915\u094d\u0937\u0930\u094b\u0902 \u0915\u093e \u0909\u092a\u092f\u094b\u0917 \u0915\u0930 \u0910\u0938\u0947 \u0915\u093f\u0924\u0928\u0947 4 \u0905\u0915\u094d\u0937\u0930\u094b\u0902 \u0915\u0947 \u0936\u092c\u094d\u0926 \u0915\u093e \u0928\u093f\u0930\u094d\u092e\u093e\u0923 \u0915\u093f\u092f\u093e \u091c\u093e \u0938\u0915\u0924\u093e \u0939\u0948 \u091c\u093f\u0938\u092e\u0947 \u0905\u0927\u093f\u0915\u0924\u092e \u0924\u0940\u0928 consonants \u0914\u0930 \u0915\u092e \u0938\u0947 \u0915\u092e 1 vowel \u0939\u094b?<\/p>\n\n\n\n A) 65<\/p>\n\n\n\n B) 360<\/p>\n\n\n\n C) 720<\/p>\n\n\n\n D) 1560<\/p>\n\n\n\n E) None of these<\/p>\n\n\n\n ANSWER KEYS and SOLUTIONS:<\/strong><\/p>\n\n\n\n Solution 1: 4)<\/strong><\/p>\n\n\n\n Desired number of ways = (8<\/sup>C5<\/sub> \u00d7 6<\/sup>C3<\/sub>) + (8<\/sup>C6<\/sub> \u00d7 6<\/sup>C2<\/sub>) + (8<\/sup>C7<\/sub> \u00d7 6<\/sup>C1<\/sub>) + 8<\/sup>C8<\/sub> = 1120 + 420 + 48 + 1 = 1589<\/p>\n\n\n\n Hence, option d.<\/p>\n\n\n\n Solution 2: 2)<\/strong><\/p>\n\n\n\n Desired number of ways {7!\/(3! \u00d7 2!)} \u00d7 {14!\/(3! \u00d7 3! \u00d7 2!)} = 12740 \u00d7 11!<\/p>\n\n\n\n Hence, option b.<\/p>\n\n\n\n Solution 3: 3)<\/strong><\/p>\n\n\n\n Total number of different numbers of 4 digits formed using these digits only once = 4! = 24<\/p>\n\n\n\n The number formed whose unit digit is \u20182\u2019 = 6<\/p>\n\n\n\n The number formed whose unit digit is \u20183\u2019 = 6<\/p>\n\n\n\n The number formed whose unit digit is \u20184\u2019 = 6<\/p>\n\n\n\n The number formed whose unit digit is \u20185\u2019 = 6<\/p>\n\n\n\n Sum of the total number of unit digits = 2 \u00d7 6 + 3 \u00d7 6 + 4 \u00d7 6 + 5 \u00d7 6 = 84<\/p>\n\n\n\n Desired sum = 84 + 84 \u00d7 10 + 84 \u00d7 100 + 84 \u00d7 1000 = 84 \u00d7 (1000 + 100 + 10 + 1) = 84 \u00d7 1111 = 93324<\/p>\n\n\n\n Hence, option c.<\/p>\n\n\n\n Solution 4: 4)<\/strong><\/p>\n\n\n\n There are 5 consonants and 3 vowels in the word \u201cCASTLIEY\u201d<\/p>\n\n\n\n Case I: 1 vowel and 3 consonants are there in the word<\/p>\n\n\n\n Number of ways of selecting letters = 3<\/sup>C1<\/sub> \u00d7 5<\/sup>C3<\/sub> = 3 \u00d7 10 = 30<\/p>\n\n\n\n Case II: 2 vowels and 2 consonants are there in the word<\/p>\n\n\n\n Number of ways of selecting letters = 3<\/sup>C2<\/sub> \u00d7 5<\/sup>C2<\/sub> = 3 \u00d7 10 = 30<\/p>\n\n\n\n Case III: 3 vowels and 1 consonant are there in the word<\/p>\n\n\n\n Number of ways of selecting letters = 3<\/sup>C3<\/sub> \u00d7 5<\/sup>C1<\/sub> = 1 \u00d7 5 = 5<\/p>\n\n\n\n So the total number of ways in which the letter are selected = 30 + 30 + 5 = 65<\/p>\n\n\n\n These 4 letters can be arranged with in the word in 4! = 24 ways<\/p>\n\n\n\n So the total number of words formed = 24 \u00d7 65 = 1560<\/p>\n\n\n\n Hence, option d.<\/p>\n\n\n\n Solution 5: 3)<\/strong><\/p>\n\n\n\n There are 4 even positions and 3 vowels.<\/p>\n\n\n\n Number of ways of filling these 3 vowels = 4<\/sup>P3<\/sub> = 24<\/p>\n\n\n\n Number of ways of filling 5 consonants = 5!\/2! = 60<\/p>\n\n\n\n So, the total number of words formed = 60 \u00d7 24 = 1440<\/p>\n\n\n\n Hence, option c.<\/p>\n\n\n\n <\/p>\n\n\n\n ALSO READ: Word Swap Tricks for RBI Grade B Phase 1 Exam<\/a><\/strong><\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n Are you preparing for the RBI Grade B exam 2025? If so, it’s the ideal time to start and accelerate your exam preparation. The notification can be released anytime soon! <\/strong><\/p>\n\n\n\n Start preparing with the course that best suits you below!<\/a><\/strong><\/p>\n\n\n\nCommon Types of Permutation and Combination Questions<\/strong><\/h2>\n\n\n\n
Simple Permutations<\/strong><\/h3>\n\n\n\n
Permutations having Repetition<\/strong><\/h3>\n\n\n\n
Restricted Permutations<\/strong><\/h3>\n\n\n\n
Simple Combinations<\/strong><\/h3>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-rt.googleusercontent.com\/docsz\/AD_4nXe_xO4XJjvqiqrjh40v3bS2zlBj8eEUmxL1_RkQ1BYguaPk7SCzhGDU6skCZtvOfravh9eQmPk5jFsXinNlAv0wBag4aqvX6zEGt8wBYfOfpwO73JeSZ5SrUkL4NsqcFdi7YyN0CQ?key=U4ZwLLFvEvlVq-kCSwzd0aDp\")
Combinations with Restrictions<\/strong><\/h3>\n\n\n\n
Mixed Problems<\/strong><\/h3>\n\n\n\n
![\"\"](\"https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2025\/08\/image-25.png\")
<\/figure>\n\n\n\n5 Important Formulas to Remember<\/h2>\n\n\n\n
1. Factorial Rule<\/h3>\n\n\n\n
2. Permutation Formula<\/h3>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-rt.googleusercontent.com\/docsz\/AD_4nXdbNQfrA4S7YpmlvDCmFEqqGna-MfWRc931g1q9428uCfUKeNu_lFLws7BFS_ADYgb_XmC4JWC7pC5kK1e4Ms-TIC8ngg8hQi9ui0H_WR8QDbRHd9ImBMx8S7i4jNMm3Q9Gjm4zOQ?key=U4ZwLLFvEvlVq-kCSwzd0aDp\")
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3. Combination Formula<\/h3>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-rt.googleusercontent.com\/docsz\/AD_4nXdzpqMH336lvzNoCz6nVwICEYbkdRWOvXMN9dFjpxXn0Ub_TdznikjxyDZ2lWLk7_4EZRigltoxuBrKF92-DlTYA4x-YJHeLb4a5JlBiY2UQdYcuoNaAgxzVfpbFNRjCd0Hj7mP?key=U4ZwLLFvEvlVq-kCSwzd0aDp\")
4. Permutations with Repetition<\/h3>\n\n\n\n
5. Principle of Fundamental Counting<\/h3>\n\n\n\n
Tips to Solve Permutation and Combination Quickly<\/h2>\n\n\n\n
Types of Questions Asked in RBI Grade B Phase 1 Exam<\/h2>\n\n\n\n
Example 1: Simple Permutation<\/h3>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-rt.googleusercontent.com\/docsz\/AD_4nXcXr7DyKoMSYJs6oiQD8RLJik-7giBJCEEkBs5qGP9i6P7n2kZTrPIyYtWTHITBk25Zo5s4M1Euf4mmDF2tga_O_Z1dmdIT_Af8wfKFdLTxNDH8gmwu0_XEovp0c55gradmjXOyuA?key=U4ZwLLFvEvlVq-kCSwzd0aDp\")
Example 2: Permutation with Restriction<\/h3>\n\n\n\n
Arrangements = 3!=63! = 6
BA can be AB as well, so multiply by 2 \u2192 Total = 12 ways.<\/p>\n\n\n\nExample 3: Simple Combination<\/h3>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-rt.googleusercontent.com\/docsz\/AD_4nXdejOMmb3Da0ATsAvXE2NyNbhTPHTIADdBOGVcqbM-zI00jLJGy2yEL0glEn1g8LudTwCczgJYhFAfZQqp8fOyblQgzI65rh9aU7f87HC4QaULPRBt1Jp6qkiTuO6DC-H3cv2qBPA?key=U4ZwLLFvEvlVq-kCSwzd0aDp\")
Example 4: Combination with Restriction<\/h3>\n\n\n\n
Shortcuts and Tricks You Must Know<\/h2>\n\n\n\n
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How to Practise Effectively<\/h2>\n\n\n\n
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Mini Task for You<\/h3>\n\n\n\n
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Sample Questions for Practice<\/h2>\n\n\n\n
1) – 4)<\/td> 2) – 2)<\/td> 3) – 3)<\/td> 4) – 4)<\/td> 5) – 3)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
![\"\"](\"https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2025\/06\/How-to-Make-RBI-Grade-B-2025-Exam-Preparation-Notes-for-Phase-1-and-Phase-2.png\")
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