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Permutation & Combination

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Permutation & Combination

  1. Permutation
  • Definition: Permutations refer to the different ways of arranging a set of objects, where the order matters.
  • Formula: The number of ways to arrange n distinct objects in r places is given by: P(n,r)=n!/(n−r)!
  • Example: If you have 3 letters A, B, and C, the number of ways to arrange all of them is P(3,3)=3!=6P (i.e., ABC, ACB, BAC, BCA, CAB, CBA).
  1. Combination
  • Definition: Combinations refer to the different ways of selecting a set of objects, where the order does not matter.
  • Formula: The number of ways to choose rrr objects from nnn objects is: C(n,r)=n!/r!(n−r)!
  • Example: If you have 3 letters A, B, and C, the number of ways to choose 2 letters is C(3,2)=3!/2!(3−2)!=3 (i.e., AB, AC, BC).

Key Points to Remember:

  • In permutations, the arrangement matters (e.g., AB and BA are different).
  • In combinations, the arrangement does not matter (e.g., AB and BA are the same).
  • Factorials n!n!n! represent the product of all positive integers up to nnn, and they are

Let’s look at some examples of Permutation and Combination that are aligned with the level of difficulty typically seen in banking exams:

Example 1: Permutation

Question: In how many different ways can the letters of the word “BANK” be arranged?

Solution:The word “BANK” consists of 4 distinct letters.

The number of ways to arrange all the letters is given by 4! (since there are no repeating letters). 4!=4×3×2×1=24

Thus, the total number of arrangements of the word “BANK” is 24.

Example 2: Combination

Question: A committee of 4 members is to be formed from a group of 7 people. In how many ways can this be done?

Solution: Here, we are selecting 4 people out of 7, and the order of selection does not matter, so this is a combination problem.

The number of ways to choose 4 members from 7 is given by C(7,4): C(7,4)=7!/4!(7−4)!=7×6×5×4!/4!×3!=7×6×5/3×2×1=35

Thus, the committee can be formed in 35 different ways.

 

 

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