Title XI - maths - chapter 7 - PERMUTATIONS (1-29).pdf ✅

NARAYANAGROUP 1 JEE-MAIN SR.MATHS-VOL-II PERMUTATIONS & COMBINATIONS PERMUTATIONS Fundamental Principle :  Multiplication Principle: If an operation can be performed in 'm' different ways and another operation in 'n' different ways then these two operations can be performed one after the other in 'mn' ways. Note: Here the key word is and, so and represents multiple rule W.E-1: In a shelf there are ‘2’ different physics books and ‘3’ different chemistry books. The number of ways in which a student can select a physics book and chemistry book is Sol. A student can select a physics book and a chemsitry book in 2 3 6   ways as explained below. Let the physics books be P1 , P2 and chemistry books be C1 , C2 , C3 . The six selections are P1 C1 , P1 C2 , P1 C3 , P2 C1 , P2 C2 , P2 C3 . W.E-2: A two wheeler agent sells scooters, motorcycles. In each body pattern two capacities 100 c.c. and 150 c.c. available. In each capacity there are five colours. The number of choices a customer will have to buy a vehicle is Sol. The number of choices a customer will have to buy a vehicle is 2 x 2 x 5 = 20.  Addition Principle: If an operation can be performed in 'm' different ways or another operation in 'n' different ways then either of these two operations can be performed in 'm + n' ways (provided only one has to be done). Note 1: Here key word is or, so or represents Addition rule Note 2:This principle can be extended to any finite number of operations. W.E-3: In a shelf there are ‘2’ different physics books and ‘3’ different chemistry books. The number of ways in which a student can select a physics book or a chemistry book is Sol. A student can select a physics book or a chemistry book in (2 + 3) = 5 ways. The five ways are P1 , P2 , C1 , C2 , C3 . W.E-4: There are 15 two bed room flats in a building and 10 two bed room flats in other building and 8 two bed room flats in a third building. The number of choices a customer will have for buying a flat is Sol. The number of choices a customer will have for buying a flat is 15 + 10 + 8 = 33. Factorial ‘n’  The continuous product of first 'n' natural numbers is called factorial n and is denoted by n! (or) n i.e., n! = 1 x 2 x 3 x .... (n-1) x n. n! = n [(n-1)!]  0! = 1 6! = 720 1! = 1 7! = 5040 2! = 2 8! = 40320 3! = 6 9! = 362880 4! = 24 10! = 3628800 5! = 120 (2n)! = 2n . n! . [1.3.5 ... (2n-1)]  Permutation: An arrangement that can be formed by taking some or all of a finite set of things (or objects) is called a Permutation. Order of the things is very important in case of permutation.  Linear Permutation: A permutation is said to be a Linear Permutation if the objects are arranged in a line. A linear permutation is simply called as a permutation. W.E.5:- From the letters of the word ‘RULE’. The number of two letter permutations is Sol. From the letters of the word ‘RULE’, two letter permutations (linear permutations) are RU, UR, RL, LR, RE, ER, UL, LU, UE, EU, LE, EL. PERMUTATIONS & COMBINATIONS SYNOPSIS
NARAYANAGROUP 2 JEE-MAIN SR.MATHS-VOL-II PERMUTATIONS & COMBINATIONS Permutations of ‘n’ Dissimilar things:  The number of (linear) permutations that can be formed by taking r things at a time from a set of n dissimilar things (r < n) is denoted by n Pr or P(n, r) or   n r p  The number of permutations of n dissimilar things taken r at a time is equal to the number of ways of filling of r blank places in a row by n dissimilar things.  n Pr = n(n-1) (n-2) .... (n-r+1) = ( )! ! n r n  n n r    The number of permutations of n dissimilar things taken all at a time is n Pn = n! = n Properties of n pr  P 1 n 0  n Pr = (n-1)Pr +r . (n-1)Pr-1 or n Pr + r . n Pr-1= (n+1)Pr  n Pr = n (n-1)P(r-1) = n (n-1) . (n-2)P(r-2) etc.  1 n r n r P P  = (n - r + 1) and ( 1) 1 n r n r P n P     If r < s < n, then n Ps is divisible by n Pr .  If 1 1 n n n P P P r r r a b c     then   2 b a b c    1 2  1 1 2 1 1. 2. .... . 1 n n P P n P P n n         Number of permutations of n different things, taken r at a time, when a particular thing is to be always included in each arrangement, is r.n-1Pr-1  Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement is n-1Pr W.E-6: The number of permutations of 6 different things taken 4 at a time, when a particular thing is to be always included in each arrangement is Sol. A particular thing is to be always included in each arrangement is 5 3 4 P W.E-7: The number of permutations of 6 different things taken 4 at a time, when a particular thing is never taken in each arrangement is Sol. A particular thing is never taken in each arrangement is 5 P4 .  Number of permutations of n different things, taken all at a time, when m specified things always come together is m!.(n - m + 1) !  Number of permutations of n different things, taken all a time when m specified things never come together is n! - [m! . (n - m + 1)!]  The number of permutations of 'n' dissimilar things taken 'r' at a time when 'k' particular things never occur is (n-k)Pr .  The number of permutations of 'n' dissimilar things taken 'r' at a time when k (NARAYANAGROUP 3 JEE-MAIN SR.MATHS-VOL-II PERMUTATIONS & COMBINATIONS  A number is divisible by 4, if the last two digits of the number formed is divisible by 4.  A number is divisible by 5, if the last digit is either 0 or 5.  A number is divisible by 6, if the number is divisible by 2 and 3. (Even numbers whose sum of digits is divisible by 3). W.E-10: The number of 4 digit numbers which are divisible by 6 by using the digits 2, 3, 5, 8. Sol. Given digits are 2, 3, 5, 8. A number is divisible by 6 means it is divisible by 2 and 3. A number is divisible by 3, if the sum of the digits is multiple of three. Here 2 + 3 + 5 + 8 = 18 = 3 x 6. A number is divisible by 2, if the units digit is even number (2, 8) (2 ways) Here unit’s place filled in 2 ways (2 or 8). Now remaining ‘3’ gaps are filled in 3 ways.  required numbers which are divisible by 6 are = 3 2 12   .  A number is divisible by 7, if the difference between twice the digit in the units place and the number formed by the other digits is either 0 or a multilple of 7. W.E-11: Is this number 3675 is divisible by 7. Sol. Let the number is 3675 Here units place digit is 5 Twice the unit’s place = 2 x 5 = 10 Now the remaining number is ‘367’ Then the difference between 367 and 10 is 357 clearly 357 is divisible by ‘7’  3675 is divisible by ‘7’.  A number is divisible by 8, if the number formed by the last three digits is divisible by 8. Example : 2192,9128  A number is divisible by 9, if the sum of its digits is divisible by 9. Example : 6453, 8640  A number is divisible by 10, if the last digit is 0.  A number is divisible by 11, if the sum of the digits in the odd places and the sum of the digits in the even places are equal or differ by a multiple of 11. Example : 209, 3564 Sum of Numbers: Sum of the numbers formed by taking all the given n digits (excluding 0) is (Sum of all the n digits) x (n-1)! x (111 ... n times).  Sum of the numbers formed by taking all the given n digits (including 0) is (sum of all the n digits)[(n- 1)! x (111 .... n times) - (n-2)! (111 .... (n-1) times)]  Sum of all the r-digit numbers formed by taking the given n digits (excluding 0) is (sum of all the n digits) x (n-1)Pr-1 x (111 ...... r times).  Sum of all the r-digit numbers formed by taking the given n digits (including 0) is (sum of all the n digits) [(n-1)Pr-1 x (111 .... r times) - (n-2)Pr-2 x (111 ... (r-1) times)]. W.E-12: Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition. Sol. Put n = 5, r = 4 5 1 4 1 (0 2 4 7 8) [ P         5 2 4 2 (1111) (111)] P    4 3 3 2    21[( (1111) (111)] P P    21[(24 (1111) 6(111)]   21[(26664 666)]   21 25998  545958  The sum of the digits in any place of all the numbers formed with the help of a1 , a2 , .... an (excluding zero) taken all at a time is n a a a     1 !. .......   1 2 n  W.E.13:- The sum of the digits at the 100’s place of all the numbers formed with the digits of 5,6,7,8 taken all at a time is Sol.The sum of the digits at the 100’s place of all the numbers formed with the digits of 5, 6, 7, 8 taken all at a time is = (n-1)! x (sum of the given digits)       (4 1)! (5 6 7 8)   3! 26 156  When n digits are given excluding zero, then the sum of the value of digits in any place of n digited number is (n - 1)! x sum of the numbers x {its place value}. W.E-14: The sum of the ‘value’ of the digits at the 100’s place of all the numbers formed with digits of 5, 6, 7, 8 taken all at a time is Sol. Sum of the value of the digits at the 100’s place of all the numbers formed with digits of 5, 6,7, 8 taken all at a time is (n-1)! x (sum of the given digits) x 100 = (4-1)! x (5 + 6 + 7 + 8) x 100 = 3! x 26 x 100 = 6 x 26 x 100 = 15600
NARAYANAGROUP 4 JEE-MAIN SR.MATHS-VOL-II PERMUTATIONS & COMBINATIONS Permutations when repetitions are allowed:  The number of permutations of n dissimilar things taken r at a time when repetition of things is allowed any number of times is nr .  The number of permutations of n things (distinct) taken r at a time with atleast one repetition = . r n r n P  W.E-15: The number of 4 digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 that are divisible by 3 when repetition is allowed Sol. Fill the first 3 places with the given 6 digits in 63 ways. 2 or 8 (6) (6) (6) (2 ways) Now, after filling up the first 3 places with three digits, if we fill up the units place in 6 ways, we get 6 consecutive positive integers. Out of any six consecutive integers exactly two are divisible by 3. Hence the units place can be filled in 2 ways. Hence the number of 4 digit numbers divisible by 3 is 2 x 63 = 432. W.E-16: The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 that are divisible by 6 when repetition is allowed. Sol. Fill the first 3 places with the given 6 digits in 63 ways. (6) (6) (6) (1 way) Now, after filling up the first 3 places with three digits, if we fill up the units place in 6 ways, we get 6 consecutive positive integers. Out of any six consecutive integers exactly one is divisible by 6. Hence the units place can be filled in one way. Hence the number of 4 digit numbers divisible by 6 is 1 x 63 = 216. W.E-17: Let A be a set of n (>3) distinct elements. The number of triplets (x, y, z) of the elements of A in which at least two coordinates are equal is Sol. Here at least two coordinates are equal. i.e two (or) three cordinates are equal.  At least one coordinate is repeated Hence r r P n n  where r  3,  3 3 P n n   The number of permutations of n different things, taken not more than r at a time, when each thing may occur any number of times = n + n2 + n3 + ....+ nr = 1 ( 1)   n n nr W.E-18: The number of positive integers having atmost 20 digits using the digits 1, 2, 3, 4, 5, 6. Sol. Required number of numbers = 6 + 62 + 63 +..........+ 620 = 20 6(6 1) 5   The number of permutations of n different things taken not more than ‘r’ at a time (without repetition) = n P1 + n P2 + n P3 + ...... + n Pr  Number of positive numbers not more than P digits, if the digits 0, 1, ....,9 are used when repetitions of digits are allowed is 1 P n  where n is number of digits including zero. W.E-19: The number of positive integers having atmost 10 digits using 0,1,2,3,4,5,6,7 Sol. Required number of numbers = 810 - 1.  The number of permutations of ‘n’ different things taken more than ‘r’ at a time when repetitions are allowed is r n   1 2 ....... r r n n n n      =   1 n r n n n n   Number of functions :  The number of mappings that can be defined from set A containing ‘m’ elements to set B containing ‘n’ elements is nm.  The number of injections (one one functions) that can be defined from set A containing ‘m’ elements to set B containing ‘n’ elements is P , n m where n m  The number of surjections (on to functions) that can be defined from set A containing ‘m’ elements to set B containing ‘2’ elements is 2 2  m  The number of bijections (one one and onto functions) that can be defined from set ‘A’ containing ‘n’ elements to set B containing ‘n’ elements is n!  The number of many to one functions that can be defined from set A containing ‘m’ elements to set B containing ‘n’ elements is  m n n Pm