Quantitative Aptitude-Concepts and Theory updated on Jul 2019
A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. Permutation and combination is all about counting and arrangements made from a certain group of data.

# Permutation and Combination-Concepts and Theory

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Permutation and Combination  Aptitude basics, practice questions, answers and explanations
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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.

A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. Permutation and combination is all about counting and arrangements made from a certain group of data. Freshersworld.com provides questions and answers with explanations on Permutation and combination. It also provides with practice questions, solved examples that would help candidates in clearing all the competitive tests. It also provides online test on Permutation and combination - quantitative Aptitude. What is a formula of permutation? n!/(n-r)! What is the difference between permutation and combination? In permutation order of things is considered, whereas in combination, order of things doesnt matter. Example of permutation: Picking first, second and third place winners. Example of combination: Picking three winners.