Permutation and Combination For IBPS RRB Exam, Check Tips and Tricks

Permutation and Combination For IBPS RRB Exam: Permutations and combinations are important topics for IBPS RRB exams. Permutations tackle arranging objects in a particular order. The order matters in permutations. Combination, on the other hand, refers to the selection of items without considering the order. The order does not matter in combinations. Combinations focus on selecting objects without considering their order.

In this blog, we’ll dive into the formulas, tips and tricks, and question samples needed to master Permutation and Combination for IBPS RRB Exam.

Permutation and Combination For IBPS RRB Exam, Check Tips and Tricks

To make the most of the questions on Permutation and Combination for IBPS RRB Exam, you need to need to understand the basics and the core of it through formulas and examples. Given below are the tips and tricks along with the formulas to tackle this topic.

Permutations Formula:

The formula for permutations of (n) objects taken (r) at a time is: [ P(n, r) = \frac{n!}{(n-r)!} ] where (n!) represents the factorial of (n).

To make it easy, For example, if you have three letters A, B, and C, the different permutations would be ABC, ACB, BAC, BCA, CAB, and CBA.

Example: If you want to arrange 3 letters out of ABCD, there are (4P3 = 24) ways.

Combinations Formula:

The formula for combinations of (n) objects taken (r) at a time is: [ C(n, r) = \frac{n!}{r! \cdot (n-r)!} ]

To make it easy, For example, if you have three letters A, B, and C, and you need to select two at a time, the combinations would be AB, AC, and BC.

Example: If you want to form a committee of 2 males and 3 females from a pool of 8 males and 9 females, there are (C(8, 2) \cdot C(9, 3) = 2352) ways.

Permutation and Combination For IBPS RRB Exam, Check Tips and Tricks

To master Permutation and Combination for IBPS RRB Exam, you’ll have to read the question properly, look for keywords, master the use of factorials, and solve a different type of problems, know how permutations and combinations relate to probability, and more. Given below is the table that encapsulates the best tips and tricks needed to master the Permutation and Combination for IBPS RRB Exam:

Permutation and Combination For IBPS RRB Exam, Check Tips and Tricks
Tip/Trick	Description
Understand the Problem	Read the question carefully to decide or ascertain whether it involves permutations or combinations.
Identify Keywords	Search for words like “arrange,” ‘’order,’’ “select,” or “group.”
Practice	Solve a variety of problems to build intuition.
Use Factorials	Get used to the factorials (e.g., 5!=5×4×3×2×15! = 5 \times 4 \times 3 \times 2 \times 15!=5×4×3×2×1).
Visualize	Draw diagrams or visualize scenarios to simplify difficult problems.
Consonants Together	Treat consonants as a single unit to solve problems like “PERMUTATION’’ while arranging letters.
Vowels Together	Similarly, treat vowels as a single unit when they need to be together.
Probability	Understand how permutations and combinations relate to probability.
Additional Practice	Find permutation and combination questions in previous year papers and watch online videos related to them.

Permutation and Combination For IBPS RRB Exam, Best Sample Questions

Candidates need to practice as much as they can to master Permutation and Combination for IBPS RRB Exam. Here are the best samples questions that will help you understand the type of questions that might pop up in the IBPS RRB exam:

Question 1: A group of nine students having at least 6 boys in it is to be selected from a group of 8 boys and 7 girls. Find total number of ways in which the group can be formed.

A) 1155

B) 1145

C) 1125

D) 1175

E) None of these

Question 2: For an exhibition, a group of 7 cars has to be chosen out of 13 cars such that a particular car is always chosen. If the cars have to be chosen one by one, then find the total number of ways of doing so.

A) 848 × 6!

B) 924 × 7!

C) 544 × 7!

D) 712 × 6!

E) None of these

Question 3: A bag contains 2 blue cubes, 3 green cubes and 4 yellow cubes. In how many ways 3 cubes can be drawn from the bag such that at least one green cube is drawn?

A) 72

B) 64

C) 56

D) 48

E) 84

Question 4: 4 boys or 5 girls are to be selected from a group of 7 boys and 9 girls. Find the total number of ways in which the selection can be made.

A) 164

B) 161

C) 132

D) 144

E) None of these

Question 5: Find the total number of ways in which the word OCCUPATION can be arranged so that all vowels are together.

A) 48000

B) 18000

C) 21600

D) 72000

E) None of these

Question 6: Find the total number of ways of arrangement of the word ACCUMULATION, so that all the vowels are together.

A) 453600

B) 682400

C) 620000

D) 540500

E) None of these

Question 7: In how many ways can the letters of word ‘BOTTLE’ be arranged?

A) 1440

B) 720

C) 240

D) 360

E) None of these

Question 8: A book shelf has 8 Hindi, 7 English and 11 Sanskrit books. In how many ways can Sourav choose 5 books from the shelf so that it consists of one Hindi, two English and two Sanskrit books?

A) 8220

B) 8600

C) 9240

D) 9660

E) None of these

Question 9: A cricket team of 11 players is to be selected from a group of 8 batsmen, 5 bowlers and 3 wicketkeepers. In how many ways a team having exactly 4 bowlers and 1 wicketkeeper is selected?

A) 168 ways

B) 300 ways

C) 420 ways

D) 568 ways

E) None of these

Question 10: A basketball team of 5 players is to be selected from a group of 7 male players and 6 female players. In how many ways a team having exactly two female players is selected?

A) 525 ways

B) 510 ways

C) 475 ways

D) 420 ways

E) None of these

Solution 1: A)

Desired number of ways = ⁸C₆ × ⁷C₃ + ⁸C₇ × ⁷C₂ + ⁸C₈ × ⁷C₁ = 980 + 168 + 7 = 1155

Hence, option a.

Solution 2: B)

Since, 1 car is always chosen, therefore, 6 cars have to be chosen out of 12 cars.

Therefore, number of ways of choosing the cars = ¹²C₆ = 12!/{6! × (12 – 6)!} = 924

Required number of ways for the exhibition of cars = 924 × 7!

Hence, option b.

Solution 3: B)

The possible cases = (1 green and 2 non green cubes) or (2 green and 1 non green cubes) or (3 green cubes)

Total number of non green cubes = 2 + 4 = 6

Also, ⁿC_r = n!/{n! ×(n – r)!}

Therefore, number of ways of selecting 1 green and 2 non-green cubes = ³C₁ × ⁶C₂ = 3 × 15 = 45 ways

Number of ways of selecting 2 green cubes and 1 non-green cube = ³C₂ × ⁶C₁ = 3 × 6 = 18 ways

Number of ways of selecting 3 green cubes = ³C₃ = 1 way

Therefore, total number of ways = 45 + 18 + 1 = 64 ways

Hence, option b.

Solution 4: B)

Desired number of ways of selection = ⁷C₄ + ⁹C₅ = 35 + 126 = 161

Hence, option b.

Solution 5: C)

Desired number of ways = (6! × 5!)/(2! × 2!) = 21600

Hence, option c.

Solution 6: A)

Desired number of ways = (7! × 6!)/(2! × 2! × 2!) = 453600

Hence, option a.

Solution 7: D)

Number of letters in the word BOTTLE = 6

T appears twice.

Required number of ways = 6!/2!

= 6 × 5 × 4 × 3 = 360

Hence, option d.

Solution 8: C)

Number of ways = ⁸C₁ × ⁷C₂ × ¹¹C₂ = 8 × 21 × 55 = 9240

Hence, option c.

Solution 9: C)

So, the required number of ways = ³C₁ × ⁵C₄ × ⁸C₆ = 3 × 5 × 28 = 420 ways

Hence, option c.

Solution 10: A)

Number of ways of selecting the team = ⁷C₃ × ⁶C₂ = 35 × 15 = 525 ways

Hence, option a.

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