Quantitative Aptitude-Concepts and Theory updated on Feb 2019

Permutation and Combination-Concepts and Theory

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Permutation and Combination  Aptitude basics, practice questions, answers and explanations 
Prepare for companies tests and interviews

 

Factorial: The continued product of first 'n' natural numbers is called the 'n factorial' and is denoted by n!. That is
n!=1*2*3*4*........*(n-1)*n

Eg: 4!=1*2*3*4

1. nPr=n!/(n-r)!
2.  nPn=n!
3.  nP1=n

1. nCr = n!/(r! (n-r)!)
2. nC1 = n
3. nC0 = 1 = nCn
4. nCr = nCn-r
5. nCr =  nPr/r!

1.The number of all permutations of n distinct items or objects taken 'r' at a time is

n(n-1)(n-2)...........(n-(r-1)) = n P r

2.The number of all permutations of n distinct objects taken all at a time is n!

3.The number of ways of selecting r items or objects from a group of n distinct items or objects is nCr = n!/(r! (n-r)!) 

4. If there are  n subjects of which  p 1 are alike of one kind;  p 2 are alike of another kind;  p 3 are alike of third kind and so on and  p r are alike of  r thkind, 

such that ( p 1 +  p 2 + ...  p r ) =  n .

5.Then, number of permutations of these n objects=   n! /  (p1 !)(p2 !).........(pr !)          

                                                                      

6.The number of all combinations of n things, taken r at a time is:

nCr = n!/(r! (n-r)!)  = n(n-1)(n-2)/r!..... to r factors

7.The number of circular arrangements of n distinct items is (n – 1)! when there is a difference  between clockwise and anticlockwise arrangements and (n – 1)!/2 when there is no difference between clockwise and anticlockwise arrangements.

8.If all possible n-digit numbers using n-distinct non-zero digits formed, the sum of all the numbers so formed is equal to (n – 1)! x [sum of the n digits] x {1111…..n times}.

 

Dividing given Items into groups:

1. The number of ways of dividing (p + q) items into two groups containing p and q items respectively is  (p + q)!/p!q!
    
2. The number of ways of dividing 2p items into two equal groups of p each is (2p)! / (p!)2 when the two groups have distinct identity and (2p)!/2!(p!)2 when the two groups do not have distinct identity.
    
3. The number of ways of dividing (p + q + r) items into three groups containing p, q and r items respectively is (p + q + r)! / p!q!r!
    
4. The number of ways of dividing 3p items into three equal groups of p each is (3p)! / (p!)3 when the three groups have distinct identity and (3p)! / 3!(p!)3 when the three groups do not have distinct identity.

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