Notes | EduRev

JEE Main Mock Test Series 2020 & Previous Year Papers

JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Main Mock Test Series 2020 & Previous Year Papers.
All you need of JEE at this link: JEE
  • 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!

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of JEE

JEE

Dynamic Test

Content Category

Related Searches

video lectures

,

Notes | EduRev

,

practice quizzes

,

past year papers

,

Sample Paper

,

Summary

,

Notes | EduRev

,

Semester Notes

,

Notes | EduRev

,

Viva Questions

,

mock tests for examination

,

study material

,

Exam

,

ppt

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

pdf

,

Free

,

Important questions

,

Extra Questions

,

Objective type Questions

,

MCQs

;