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Revision Notes - Permutation Notes | Study JEE Main & Advanced Mock Test Series - JEE

Document Description: Revision Notes - Permutation for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The notes and questions for Revision Notes - Permutation have been prepared according to the JEE exam syllabus. Information about Revision Notes - Permutation covers topics like and Revision Notes - Permutation Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Revision Notes - Permutation.

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• The concept of permutation is used for the arrangement of objects in a specific order i.e. whenever the order is important, permutation is used.
• The total number of permutations on a set of n objects is given by n! and is denoted as nPn = n!
• The total number of permutations on a set of n objects taken r at a time is given by nPr = n!/ (n-r)!
• The number of ways of arranging n objects of which r are the same is given by n!/ r!
• If we wish to arrange a total of n objects, out of which ‘p’ are of one type, q of second type are alike, and r of a third kind are same, then such a computation is done as n!/p!q!r!
• Al most all permutation questions involve putting things in order from a line where the order matters. For example ABC is a different permutation to ACB.
• The number of permutations of n distinct objects when a particular object is not to be considered in the arrangement is given by n-1Pr
• The number of permutations of n distinct objects when a specific object is to be always included in the arrangement is given by r.n-1Pr-1.
• If we need to compute the number of permutations of n different objects, out of which r have to be selected and each object has the probability of occurring once, twice or thrice… up to r times in any arrangement is given by (n)r.
• Circular permutation is used when some arrangement is to be made in the form of a ring or circle.
• When ‘n’ different or unlike objects are to be arranged in a ring in such a way that the clockwise and anticlockwise arrangements are different, then the number of such arrangements is given by (n – 1)!
• If n persons are to be seated around a round table in such a way that no person has similar neighbor then it is given as ½ (n – 1)!
• The number of necklaces formed with n beads of different colors = ½ (n – 1)!
• nP0 =1
• nP1 = n
• nPn = n!/(n-n)! = n! /0! = n! /1= n!

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