Tips And Tricks And Shortcuts For Permutation And Combination

July 16, 2020

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Tips and Tricks & Shortcuts for Permutation & Combination

  • Permutation: The different arrangements of a given number of things by taking some or all at a time, are called permutations. This is denoted by nPr.
  • Combination: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination. This is denoted by nCr.

Here, are rapid and easy tips and tricks and shortcuts on Permutation and Combination questions swiftly, easily, and efficiently in competitive exams and recruitment exams.

Permutation and Combinations Tips and Tricks and Shortcuts

    • Use permutations if a problem calls for the number of arrangements of objects and different orders are to be counted.
    • Use combinations if a problem calls for the number of ways of selecting objects and the order of selection is not to be counted.
    • Summary of formula to use

In how many ways can a word be arranged.

Tricks and Tips on type 1 Question

This is Permutation Question.

Let us take this ahead as an example – 

In how many ways can the letters of the word ‘LEADER’ be arranged?

Count number of Occurances

  • L – 1
  • E – 2
  • A – 1
  • D – 1
  • R – 1

Total Unique Occurrences – 6(as E repeated 2 times)

Direct Formula = (Unique Occurrences)!/(Each Individual Unique Occurrences)

so = 6!/(1!)(2!)(1!)(1!)(1!) = 360

Type 1: Different ways to arrange (with repetition)

Question 1 In how many ways can the letters of the word ‘LEADER’ be arranged?

Options:

      1. 720
      2. 360
      3. 200
      4. 120

Solution      Letter ‘E’ appears twice and all other letters 1L, 1A, 1D and 1R appears once in the word.

Required number of ways = 6!2! \frac{6!}{2!} =6×5×4×3×2×12×1=7202 \frac{ 6 × 5 × 4 × 3 × 2 × 1}{2 × 1} = \frac{720}{2} = 360

Correct option: 2

Type 2: Different ways to arrange (without repetition)

Question 1 How many different ways are there to arrange your first three classes if they are Math, English, and Hindi?

Options:

      1. 4
      2. 6
      3. 120
      4. 36

Solution      We know that,

Pr = n!

P3 = 3!

P3 = 6

Correct option: 2

Type 3: Different ways to select (with repetition)

Question 1 In a shop there are 4 types of sweets. In how many ways can Shekhar buy 19 sweets?

Options:

      1. 480
      2. 540
      3. 720
      4. 1540

Solution      r + n – 1Cr = 19 + 4 – 1C19 =22C19

We know that, nCr = n!(n−r)!r!\frac{n!}{(n-r)! r! }

22C19 =22!(22−19)!19!\frac{22!}{(22-19)! 19! } = 1540

Type 4: Different ways to select (without repetition)

Question 1 How many different 4 digit numbers can be formed using the digits 2,3,4,5,6,7,8 no digit being repeated in any number

Options:

      1. 720
      2. 120
      3. 24
      4. 840

Solution:    The thousand place can be filled in 7 ways, the hundredth place can be filled in 6 ways, the tens place can be filled in 5 ways, and the ones place can be filled in 5 ways.

Total ways = 7*6*5*4 = 840

Correct option: 4

 

Read Also – Click for More on Permutation on Combination.

4 comments on “Tips And Tricks And Shortcuts For Permutation And Combination”


  • Shravana

    Note clear with type 3 ques.I also think it is 19^4(19X19X19X19ways)

    16
    Reply ↓

  • ARITRA

    Type 3:
    There are Four types of sweets and there are 19 places, I think the answer would have been 19^4.

    25
    Reply ↓