Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

Quantitative Ability for SSC CHSL

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Quant : Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

The document Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev is a part of the Quant Course Quantitative Ability for SSC CHSL.
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Permutation

Permutation is basically called as a arrangement where order does matters.Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time.It is represented as nPr 

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

1. nPr = n! / (n-r)!

2. If from the total set of n numbers p is of one kind and q ,r are others respectively then nPr = n! / p! × q! × r!.

3. nPn = n!

Combination

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

  1.  nCr = n!/ r! × (n-r)!  
  2.  nC0 = 1
  3.  nCn = 1
  4.  nCr =  nCn - r 
  5.  nCa =  nCb => a = b => a+b = n
  6. nC0 + nC1nC2nC3+ ...............+ nC= 2n

Permutation vs Combination

In both the things main difference is of order .In permutation order matters while in combination it does not.

Basic Difference :

  1. order
  2. arrange or choose 
  3. number of permutation > number of combination

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

In Arrangements we have,

Total no. of arrangements = total no. of groups or selection × r!

where r is the no. of objects in each group or selection. So nPr = nCr  × r!

Questions :

1. How many triangles can be formed with four points (A,B,C & D) in a plane ? It is given that no three points are col-linear(not comes in straight line).From the three points A,B and C have only one triangle with these points.

Sol:

Here in this question  , Order of digit does not matter so it is a combination.

 nCr = n!/ r! × (n-r)!  

 4C= 4!/3! × ( 4- 3)! 

 4C= 4!/3! × 1! 

 4C= 4!/3! × 1!

 4C= 4

Or

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev
Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

2. How many number plates of 3 digit can be formed with four digits 1,2,3 and 4 ?

Sol:

Here, the order of arrangement of digits does matter.

nPr = n! / (n-r)!

nPr = 4! / (4-3)!

4P3 = 4! / 1!

 4P3 = 4!

 4P3 = 24

Factorial Notation

To solve problem like this you must have the knowledge of factorial.Factorial is represented as like " ! ".The Factorial notation is :

Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 1) Quant Notes | EduRev

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