Tips And Tricks And Shortcuts For Permutation And Combination

Tips and Tricks, and Shortcuts for Permutation and Combination

Tips and Tricks for Permutation and Combination has been discussed on this page to help student practice shortcuts while solving questions.

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.

Tips and Tricks, and Shortcut for Permutation and Combination

Tips and Tricks and Shortcuts for Permutation and Combination

    • 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.
OrderRepetitionFormula
PermutationYesnr
PermutationNonpr
CombinationYesr + n – 1Cr
CombinationNonCr

Types of Tips and Tricks and Shortcuts for Permutation and Combination

Questions on Types of Tips and Tricks and Shortcuts for Permutation and Combination

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 = \frac{6!}{2!} = \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 = \frac{n!}{(n-r)! r! }

22C19 =\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

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