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Permutations and Combinations, , PHOTON INSTITUTE UJJJAIN 8890244575/8502976721, , Permutations, , (1) Arranging n objects, taken r at a time, equivalent to filling r places from n things., , Introduction, (1) The Factorial : Factorial notation: Let n be a, positive integer. Then, the continued product of first n, natural numbers is called factorial n, to be denoted by n !, or n ., Also, we define 0 ! = 1., when n is negative or a fraction, n ! is not defined., Thus, n ! = n (n – 1) (n – 2) ......3.2.1., (2) Exponent of Prime p in n ! : Let p be a prime, number and n be a positive integer. Then the last, integer amongst 1, 2, 3, .......(n – 1), n which is divisible, n , n , by p is   p , where   denotes the greatest integer, p, p,  , less than or equal to, , n, ., p, , Definition of permutation, The ways of arranging or selecting a smaller or an, equal number of persons or objects at a time from a, given group of persons or objects with due regard being, paid to the order of arrangement or selection are called, the (different) permutations., For example : Three different things a, b and c are, given, then different arrangements which can be made, by taking two things from three given things are ab, ac,, bc, ba, ca, cb., Therefore the number of permutations will be 6., , Number of permutations without repetition, , r-places : 1, , 2, , 3, , r, , 4, , Number of choices : n (n – 1) (n – 2) (n – 3), , n – (r – 1), , The number of ways of arranging = The number, of ways of filling r places., = n(n  1)(n  2).......( n  r  1), =, , n(n  1)(n  2).....( n  r  1)((n  r)!), n!, ,  n Pr, (n  r)!, (n  r)!, , (2) The number of arrangements of n different, objects taken all at a time = n Pn  n!, (i), , n, , P0 , , n!,  1; n Pr  n . n 1 Pr 1, n!, , (ii) 0 !  1;, , 1,  0 or (r)!  (r  N ), (r)!, , Number of permutations with repetition, (1) The number of permutations (arrangements), of n different objects, taken r at a time, when each, object may occur once, twice, thrice,........upto r times, in any arrangement = The number of ways of filling r, places where each place can be filled by any one of n, r, 4, 1, 2, 3, objects., r – places :, n, n, n, n, n, Number of choices :, The number of permutations = The number of, ways of filling r places = (n)r ., (2) The number of arrangements that can be, formed using n objects out of which p are identical (and, of one kind) q are identical (and of another kind), r are

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Permutations and Combinations, identical (and of another kind) and the rest are distinct, n!, is, ., p! q!r!, , Conditional permutations, (1) Number of permutations of n dissimilar things, taken r at a time when p particular things always occur, = n  p C r  p r! ., (2) Number of permutations of n dissimilar things, taken r at a time when p particular things never occur, = n p C r r! ., (3) The total number of permutations of n, different things taken not more than r at a time, when, each thing may be repeated any number of times, is, n(n r  1), ., n 1, (4) Number of permutations of n different things,, taken all at a time, when m specified things always, come together is m ! (n  m  1)! ., (5) Number of permutations of n different things,, taken all at a time, when m specified things never come, together is n !m !  (n  m  1)! ., (6) Let there be n objects, of which m objects are, alike of one kind, and the remaining (n  m ) objects are, alike of another kind. Then, the total number of mutually, distinguishable permutations that can be formed from, n!, these objects is, ., (m !)  (n  m )!, The above theorem can be extended further i.e., if, there are n objects, of which p1 are alike of one kind;, , p 2 are alike of another kind; p 3 are alike of 3rd, kind;......; p r, are alike of rth kind such that, , arrangement of beads in a necklace, arrangement of, flowers in garland etc. then the number of circular, (n  1) !, permutations of n distinct items is, ., 2, (i) Number of circular permutations of n different, things, taken r at a time, when clockwise and, n, P, anticlockwise orders are taken as different is r ., r, (ii) Number of circular permutations of n, different things, taken r at a time, when clockwise and, n, Pr, anticlockwise orders are not different is, ., 2r, Theorems on circular permutations, Theorem (i) : The number of circular, permutations of n different objects is (n  1)! ., Theorem (ii) : The number of ways in which n, persons can be seated round a table is (n  1)! ., Theorem (iii) : The number of ways in which n, different beads can be arranged to form a necklace, is, 1, (n  1)! ., 2, , Combinations, Definition, Each of the different groups or selections which, can be formed by taking some or all of a number of, objects, irrespective of their arrangements, is called a, combination., Notation : The number of all combinations of n, things, taken r at a time is denoted by C (n, r) or, n, , n, , p1  p 2  ......  p r  n ; then the number of permutations, of these n objects is, , n!, ., ( p 1 !)  ( p 2 !)  ......  ( p r !), , Circular permutations, In circular permutations, what really matters is, the position of an object relative to the others., Thus, in circular permutations, we fix the position, of the one of the objects and then arrange the other, objects in all possible ways., There are two types of circular permutations :, (i) The circular permutations in which clockwise, and the anticlockwise arrangements give rise to, different permutations, e.g. Seating arrangements of, persons round a table., (ii) The circular permutations in which clockwise, and the anticlockwise arrangements give rise to same, permutations, e.g. arranging some beads to form a, necklace., Difference between clockwise and anticlockwise arrangement : If anti-clockwise and, clockwise order of arrangement are not distinct e.g.,, , n, C r or   ., r, C r is always a natural number., , Difference, between, a, permutation, and, combination :, (i) In a combination only selection is made whereas, in a permutation not only a selection is made but also, an arrangement in a definite order is considered., (ii) Each combination corresponds to many, permutations. For example, the six permutations ABC,, ACB, BCA, BAC, CBA and CAB correspond to the same, combination ABC., , Number of combinations without repetition, The number of combinations (selections or, groups) that can be formed from n different objects, n!, taken r(0  r  n) at a time is n C r , . Also, r !(n  r)!, n, , Cr  n Cn  r ., , Let the total number of selections (or groups) = x., Each group contains r objects, which can be arranged in, r ! ways. Hence the number of arrangements of r

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Permutations and Combinations, , PHOTON INSTITUTE UJJJAIN 8890244575/8502976721, , objects = x (r!) . But the number of arrangements =, n, , Pr ., ,  x (r!)  n Pr  x , , n, , Pr, n!, x, n Cr ., r!, r !(n  r) !, , Number of combinations with repetition and all, possible selections, (1) The number of combinations of n distinct, objects taken r at a time when any object may be, repeated any number of times., = Coefficient of x r in (1  x  x 2  .......  x r )n, = Coefficient of x r in (1  x ) n  n r 1 C r, (2) The total number of ways in which it is, possible to form groups by taking some or all of n, things at a time is n C1  n C 2  ........  n Cn  2n  1 ., (3) The total number of ways in which it is, possible to make groups by taking some or all out of, n  (n1  n 2  ....) things, when n1 are alike of one kind,, n 2 are alike of second kind, and so on is, , {(n1  1)(n 2  1)......}  1 ., (4) The number of selections of r objects out of n, identical objects is 1., (5) Total number of selections of zero or more, objects from n identical objects is n  1 ., (6) The number of selections taking at least one, out of a1  a2  a3  ......  an + k objects, where a1 are, , a 2 are alike (of second kind) and, , alike (of one kind),, , so on...... a n are alike (of nth kind) and k are distinct, = [(a1  1)(a 2  1)(a 3  1).......( an  1)] 2 k  1 ., , Conditional combinations, (1) The number of ways in which r objects can be, selected from n different objects if k particular objects, are, (i) Always included =, included =, , n k, , n k, , C r k, , (ii), , Never, , Cr, , (2) The number of combinations of n objects, of, which p are identical, taken r at a time is, n p, , C r  n  p C r 1  n  p C r  2  .......  n  p C 0 , if r  p and, , n p, , Cr  n p Cr 1  n p Cr 2  .......  n p Cr  p , if r  p ., , Division into groups, Case I : (1) The number of ways in which n, different things can be arranged into r different groups, is n  r 1 Pn or n ! n 1 C r 1 according as blank group are or, are not admissible., (2) The number of ways in which n different, things can be distributed into r different group is, r  C1 (r  1)  C 2 (r  2)  .........  (1), n, , r, , Coefficient of x, , n, , n, , r, , n, , is n ! (e  1) ., x, , r, , n 1 n, , C r 1 or, , Here blank groups are not allowed., (3) Number of ways in which m × n different, objects can be distributed equally among n persons (or, numbered groups) = (number of ways of dividing into, (mn )!n ! (mn )!, groups) × (number of groups) ! =, ., , (m !)n n! (m !)n, Case II : (1) The number of ways in which (m  n), different things can be divided into two groups which, contain, m, and, n, things, respectively, is,, (m  n)!, m n, n, Cm . Cn , ,m  n ., m !n!, Corollary: If m  n , then the groups are equal, size. Division of these groups can be given by two, types., Type I : If order of group is not important : The, number of ways in which 2n different things can be, (2n)!, divided equally into two groups is, ., 2!(n!) 2, Type II : If order of group is important : The, number of ways in which 2n different things can be, divided equally into two distinct groups is, (2n)!, 2n!,  2! , ., 2!(n!)2, (n!)2, (2) The number of ways in which (m + n + p), different things can be divided into three groups which, contain m, n and p things respectively is, (m  n  p)!, m n  p, Cm .n p Cn . p C p , ,m  n  p ., m !n ! p !, Corollary : If m  n  p , then the groups are equal, size. Division of these groups can be given by two, types., Type I : If order of group is not important : The, number of ways in which 3p different things can be, (3 p ) !, divided equally into three groups is, ., 3!( p!) 3, Type II : If order of group is important : The, number of ways in which 3p different things can be, divided equally into three distinct groups is, (3 p)!, (3 p) !, 3! , ., 3, 3!( p!), ( p!)3, (i) If order of group is not important : The, number of ways in which mn different things can be, mn !, divided equally into m groups is, ., (n!)m m !, (ii) If order of group is important: The number of, ways in which mn different things can be divided, (mn )!, (mn )!,  m! , equally into m distinct groups is, ., (n!)m m!, (n!)m, , Derangement, Any change in the given order of the things is, called a derangement., If n things form an arrangement in a row, the, number of ways in which they can be deranged so that

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Permutations and Combinations, no one of them occupies its, , 1 1 1, 1, n ! 1     ......  (1)n .  ., 1, !, 2, !, 3, !, n, !, , , original, , place, , is, , x 1  0, x 2  0, ...... x r  0, , (1) Number of total different straight lines formed, by joining the n points on a plane of which m (< n) are, collinear is n C 2  m C 2  1 ., (2) Number of total triangles formed by joining, the n points on a plane of which m (< n) are collinear is, n, C3 m C3 ., (3) Number of diagonals in a polygon of n sides is, C2  n ., , (4) If m parallel lines in a plane are intersected by, a family of other n parallel lines. Then total number of, mn (m  1)(n  1), parallelograms so formed is m C 2 n C 2 i.e.,, ., 4, (5) Given n points on the circumference of a, circle, then, , This is also equal to the coefficient of x n in the, expansion of (x 0  x 1  x 2  x 3  ......) r,  1 , = coefficient of x n in , , 1  x , , r, , = coefficient of x n in (1  x )r, = coefficient of, , x n in, , , r(r  1) 2, r(r  1)(r  2)....( r  n  1) n, x  .... , x  ......, 1  rx , 2, !, n!, , , , , r(r  1) (r  2)....( r  n  1) (r  n  1)! n r 1, , , C r 1 ., n!, n!(r  1)!, , (2) The number of integral solutions of, x1  x 2  x 3  .....  x r  n where x 1  1, x 2  1,....... x r  1 is, , (ii) Number of triangles = n C 3, (iii) Number of quadrilaterals = n C 4 ., , (x 1  x 2  x 3  ......) r ., , (6) If n straight lines are drawn in the plane such, that no two lines are parallel and no three lines are, concurrent. Then the number of part into which these, lines divide the plane is = 1  n ., (7) Number of rectangles of any size in a square, n, , of n  n is, , , , r 3 and number of squares of any size is, , r 1, , n, , 2, , where, , same as the number of ways to distribute n identical, things among r persons each getting at least 1. This also, equal to the coefficient of x n in the expansion of, , (i) Number of straight lines = n C 2, , r, , x 1  x 2  x 3  ......  x r  n, , is the same as the number of, , ways to distribute n identical things among r persons., , Some important results for geometrical problems, , n, , integral solutions of, , ., , r, , = coefficient of x n in x r (1  x )r, = coefficient of x n in, , , r (r  1) 2, r (r  1)(r  2)...(r  n  1) n, x r 1  rx , x  ... , x  .. , 2, !, n, !, , , , = coefficient of x n r in, , r 1, , (8) In a rectangle of, , n  p (n  p), , number of, , np, (n  1) ( p  1) and number of, rectangles of any size is, 4, n, , squares of any size is, ,  x , = coefficient of x n in , , 1  x , , (n  1  r)(p  1  r) ., r 1, , Multinomial theorem, Let x 1 , x 2 , ......., x m be integers. Then number of, solutions to the equation x 1  x 2  ......  x m  n, Subject to the condition, a1  x 1  b1 , a2  x 2  b 2 ,......., am  x m  bm, , .....(i), .....(ii), , n, , is equal to the coefficient of x in, (x a1  x a1 1  ......  x b1 ) (x a2  x a2 1  .....  x b 2 )......( x am  x am 1  .....  x bm ), , .....(iii), This is because the number of ways, in which sum, of m integers in (i) equals n, is the same as the number, of times x n comes in (iii)., (1) Use of solution of linear equation and, coefficient of a power in expansions to find the, number of ways of distribution : (i) The number of, , , , r(r  1) 2, r(r  1)(r  2)....( r  n  1) n, x  ... , x  ..... , 1  rx , 2!, n!, , , , =, , r(r  1)(r  2)......( r  n  r  1), r(r  1)(r  2).....( n  1), =, (n  r)!, (n  r)!, , =, , (n  1)!,  n1 Cr 1 ., (n  r)!(r  1)!, , Number of divisors, , , , , Let N  p 1 1 . p 2 2 . p 3 3 ...... p k k , where p1 , p 2 , p 3 ,...... p k, , are different primes and 1 ,  2 ,  3 ,......,  k are natural, numbers then :, (1) The total number of divisors of N including 1, and N is = (1  1)( 2  1)( 3  1)....( k  1) ., (2) The total number of divisors of N excluding 1, and N is = (1  1)( 2  1)( 3  1).....( k  1)  2 ., (3) The total number of divisors of N excluding 1, or N is = (1  1)( 2  1)( 3  1).....( k  1)  1 ., (4) The sum of these divisors is

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Permutations and Combinations, , PHOTON INSTITUTE UJJJAIN 8890244575/8502976721, ,  ( p10  p11  p12  ......  p11 ) ( p 20  p 12  p 22  ...  p2 2 )....., , taken all at a time n Cn , , ( pk0  pk1  pk2  ....  pk k ), , (5) The number of ways in which N can be, resolved as a product of two factors is,  1,  2 ( 1  1) ( 2  1)....( k  1), If N is not a perfect square, 1,  [( 1  1) ( 2  1).....(  k  1)  1], If N is a perfect square, 2, , (6) The number of ways in which a composite, number N can be resolved into two factors which are, relatively prime (or co-prime) to each other is equal to, 2 n 1 where n is the number of different factors in N., , , , n, , C0 n Cn  1, n C1  n ., , , , n, , C r  n C r 1  n  1 C r ., , , , n, , Cx  Cy  x  y or x  y  n ., , , , n . n 1Cr 1  (n  r  1)n Cr 1 ., , , , If n is even then the greatest value of, n, , , , n, , C r is, , Cn / 2 ., n, , C r is n C n 1, 2, , or C n 1 ., , , , Cr , , n n 1, . Cr 1 ., r, , Number of selections of zero or more things out, , of n different things is, n C0  n C1  n C 2  .....  n Cn  2 n .,  The number of ways of answering one or more, questions when each question has an alternative is,, 3n  1 ., ,  The number of ways of answering all of n, questions when each question has an alternative is, , P3 ., ,  The number of ways in which m things of one, type and n things of another type can be arranged in, the form of a garland so that all the second type of, m !n!, things come together , and no two things of, 2, (m  1)! m Pn, ., 2, ,  All the numbers whose last digit is an even, number 0, 2, 4, 6 or 8 are divisible by 2.,  All the numbers sum of whose digits are divisible, by 3,is divisible by 3 e.g., 534. Sum of the digits is 12,, which are divisible by 3, and hence 534 is also, divisible by 3.,  All those numbers whose last two-digit number is, divisible by 4 are divisible by 4 e.g., 7312, 8936, are, such that 12, 36 are divisible by 4 and hence the given, numbers are also divisible by 4., , 2n ., , C0  n C 2  n C4  ...... n C1  n C3  n C5  .....  2 n 1 ., , , , n, , , , 2 n 1, , , , n, , , , 6, ,  Together : Suppose we have to arrange 5, persons in a row which can be done in 5 ! = 120, ways. But if two particular persons are to be, together always, then we tie these two particular, persons with a string. Thus we have 5 – 2 + 1 (1, corresponding to these two together) = 3 +1 = 4, units, which can be arranged in 4! ways. Now we, loosen the string and these two particular can be, arranged in 2! ways. Thus total arrangements = 24 ×, 2 = 48., , second type come together , , 2, n, , these 6 gaps. Hence the answer will be, ,  The number of ways in which n (one type of, different) things and n (another type of different), things can be arranged in a row alternatively is, 2.n!.n! ., , n, , , ,  Gap method : Suppose 5 males A, B, C, D, E are, arranged in a row as × A × B × C × D × E ×. There, will be six gaps between these five. Four in between, and two at either end. Now if three females P, Q,R, are to be arranged so that no two are together we, shall use gap method i.e., arrange them in between, , Never together = Total – Together = 120 – 48 = 72., , n, , If n is odd then the greatest value of, , n!, 1, ,  1 , (0 !  1) ., n !(n  n) ! 0 !, , C0  2 n 1C1  2 n 1C 2  .....  2 n 1Cn  2 2 n ., , Cn  n 1Cn n  2 Cn n  3 Cn  ....  2 n 1Cn  2 n Cn 1 ., , Number of combinations of n dissimilar things, ,  All those numbers, which have either 0 or 5 as, the last digit, are divisible by 5.,  All those numbers, which are divisible by 2 and 3, simultaneously, are divisible by 6. e.g., 108, 756 etc.,  All those numbers whose last three-digit number, is divisible by 8 are divisible by 8.

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Permutations and Combinations, , , (c) 4, , All those numbers sum of whose digit is divisible, 7., , by 9 are divisible by 9.,  All those numbers whose last two digits are, divisible by 25 are divisible by 25 e.g., 73125, 2400, etc., , all, , numbers, , formed without repetition is, (n  1)!(a1  a2  a3  ......  an ) . Sum of the total numbers, , in this case can be obtained by applying the formula, (n  1)!(a1  a2  a3  ......  an ) . (1111......... n times)., , n, , (d) 8, , P5  20 . nP3 , then n , , (a) 4, (c) 6, , (b) 8, (d) 7, , 8., , How many words comprising of any three letters, of the word UNIVERSAL can be formed, (a) 504, (b) 405, (c) 540, (d) 450, , 9., , If, ,  If we are given n different digits (a,, a2 , a3 .......... an ) then sum of the digits in the unit place, of, , If, , n, , P4 : nP5  1 : 2 , then n  [MP PET 1987; RPET 1996], , (a) 4, , (b) 5, , (c) 6, 10., , (d) 7, In how many ways can mn letters be posted in n, letter-boxes, (b) m mn, , (a) (mn)n, 11., , Definition of permutation, Number of permutations, with or without repetition, Conditional permutations, 1., , 2., , 3., , If the best and the worst paper never appear, together, then six examination papers can be, arranged in how many ways, (a) 120, (b) 480, (c) 240, (d) None of these, How many numbers divisible by 5 and lying, between 3000 and 4000 can be formed from the, digits 1, 2, 3, 4, 5, 6 (repetition is not allowed), (a), , 6, , P2, , (b), , 5, , P2, , (c), , 4, , P2, , (d), , 6, , P3, , The number of ways in which 6 rings can be worn, on the four fingers of one hand is, (a) 4 6, (c) 6, , 4., , 5., , 4, , 6, , 13., , 14., , (a), , 4, , P4, , (b), , 4, , P3, , (c), , 4, , P1  4 P2  4 P3, , (d), , 4, , P1  4 P2  4 P3  4 P4, , 7, , C3, , 15., , 16., , 17., , 4 buses runs between Bhopal and Gwalior. If a, man goes from Gwalior to Bhopal by a bus and, comes back to Gwalior by another bus, then the, total possible ways are, (a) 12, (b) 16, , (b) 8, (d) 9, , n 1, , Pr  r, , n, , n 1, , Pr is equal to [IIT 1971; MP PET 1993], , Pr 1, , (b) n., (d), , n 1, , n 1, , Pr  n 1 Pr 1, , Pr 1  n 1 Pr, , Find the total number of 9 digit numbers which, have all the digits different, [AMU 1983], (a) 9  9 !, (b) 9 !, (c) 10 !, (d) None of these, Four dice (six faced) are rolled. The number of, possible outcomes in which at least one die shows, 2 is, (a) 1296, (b) 625, (c) 671, (d) None of these, There are 4 parcels and 5 post-offices. In how, many different ways the registration of parcel can, be made, [MP PET 1983], , (a) 20, , (b) 3 7, (d) None of these, , P5  9  n 1 P4 , then the value of n is, , (c) n(n 1 Pr  n 1 Pr 1 ), , There are 3 candidates for a post and one is to be, selected by the votes of 7 men. The number of, ways in which votes can be given is, , (c), , n, , The value of, (a), , C4, , (d) None of these, , If, , (a) 6, (c) 5, , How many numbers can be formed from the digits, 1, 2, 3, 4 when the repetition is not allowed, , (a) 7 3, , 6., , (b), , 12., , (c) n mn, (d) None of these, In how many ways can 10 true-false questions be, replied, (a) 20, (b) 100, (c) 512, (d) 1024, How many even numbers of 3 different digits can, be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, (repetition is not allowed), (a) 224, (b) 280, (c) 324, (d) None of these, , 18., , (b) 4, , 5, , (c) 5 4, (d) 5 4  4 5, In how many ways can 5 prizes be distributed, among four students when every student can take, one or more prizes, [BIT Ranchi 1990; RPET 1988, 97], , (a) 1024, (c) 120, , (b) 625, (d) 600, , [

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Permutations and Combinations, , 19., , 20., , 21., , PHOTON INSTITUTE UJJJAIN 8890244575/8502976721, , 30., In a train five seats are vacant, then how many, ways can three passengers sit [RPET 1985; MP PET 2003], (a) 20, (b) 30, (c) 10, (d) 60, The product of any r consecutive natural numbers, is always divisible by, [IIT 1985], 31., (a) r !, (b) r 2, , How many numbers of five digits can be formed, from the numbers 2, 0, 4, 3, 8 when repetition of, digits is not allowed, , (c) r n, (d) None of these, The sum of the digits in the unit place of all, numbers formed with the help of 3, 4, 5, 6 taken, all at a time is, , (a) 5, (b) 4, (c) 3, (d) 2, Assuming that no two consecutive digits are same,, the number of n digit numbers, is, (a) n!, (b) 9 !, , 32., , [Pb. CET 1990], , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , (a) 18, (b) 432, (c) 108, (d) 144, Six identical coins are arranged in a row. The, number of ways in which the number of tails is, equal to the number of heads is, (a) 20, (b) 9, (c) 120, (d) 40, The figures 4, 5, 6, 7, 8 are written in every, possible order. The number of numbers greater, than 56000 is, (a) 72, (b) 96, (c) 90, (d) 98, In how many ways can 10 balls be divided, between two boys, one receiving two and the, other eight balls, (a) 45, (b) 75, (c) 90, (d) None of these, The sum of all 4 digit numbers that can be formed, by using the digits 2, 4, 6, 8 (repetition of digits, not allowed) is, (a) 133320, (b) 533280, (c) 53328, (d) None of these, There are 5 roads leading to a town from a, village. The number of different ways in which a, villager can go to the town and return back, is, (a) 25, (b) 20, (c) 10, (d) 5, In how many ways can five examination papers be, arranged so that physics and chemistry papers, never come together, (a) 31, (b) 48, (c) 60, (d) 72, The number of ways in which first, second and, third prizes can be given to 5 competitors is, (a) 10, (b) 60, (c) 15, (d) 125, The number of 3 digit odd numbers, that can be, formed by using the digits 1, 2, 3, 4, 5, 6 when the, repetition is allowed, is, [Pb. CET 1999], (a) 60, (b) 108, (c) 36, (d) 30, , 33., , 34., , 35., , [MP PET 2000; Pb. CET 2001], , (a) 96, (c) 144, If, , 12, , Pr  1320 , then r is equal to, , 37., , 38., , 39., , (c) 7 !, (d) None of these, In how many ways n books can be arranged in a, row so that two specified books are not together, (a) n !  (n  2)!, (b) (n  1)!(n  2), , (c), , 41., , (d) (n  2) n!, , How many numbers lying between 500 and 600, can be formed with the help of the digits 1, 2, 3, 4,, 5, 6 when the digits are not to be repeated, (a) 20, (b) 40, (c) 60, (d) 80, Numbers greater than 1000 but not greater than, 4000 which can be formed with the digits 0, 1, 2,, 3, 4 (repetition of digits is allowed), are[IIT 1976; AIEEE 2, (a) 350, (b) 375, [MP PET 1996], (c) 450, (d) 576, The number of numbers that can be formed with, the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd, digits always occupy odd places, is, [RPET 1988, 1991, 19, (a) 24, (b) 18, (c) 12, (d) 30, In how many ways can 5 boys and 3 girls sit in a, row so that no two girls are together, (a) 5 !  3 !, , 40., , [, , (c) 9 n, (d) n9, The numbers of arrangements of the letters of the, word SALOON, if the two O's do not come, together, is, (a) 360, (b) 720, (c) 240, (d) 120, The number of words which can be formed from, the letters of the word MAXIMUM, if two, consonants cannot occur together, is, (a) 4 !, (b) 3 !  4 !, , (c) n !2(n  1), 36., , (b) 120, (d) 14, , 6, , P3  5 !, , (b), , 4, , P3  5 !, , (d), , 5, , P3  3 !, , How many numbers less than 1000 can be made, from the digits 1, 2, 3, 4, 5, 6 (repetition is not, allowed), (a) 156, (b) 160, (c) 150, (d) None of these, How many words can be formed from the letters of, the word COURTESY, whose first letter is C and the, last letter is Y

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Permutations and Combinations, , 42., , 43., , 44., , 45., , 46., , (a) 6 !, , (b) 8 !, , (c) 2(6) !, , (d) 2(7) !, , which are divisible by 5 while repetition of any, digit is not allowed in any number, , How many words can be made from the letters of, the word DELHI, if L comes in the middle in every, word, (a) 12, (b) 24, (c) 60, (d) 6, How many numbers consisting of 5 digits can be, formed in which the digits 3, 4 and 7 are used, only once and the digit 5 is used twice, (a) 30, (b) 60, (c) 45, (d) 90, In how many ways 3 letters can be posted in 4, letter-boxes, if all the letters are not posted in the, same letter-box, (a) 63, (b) 60, (c) 77, (d) 81, The number of 5 digit telephone numbers having, at least one of their digits repeated is, (a) 90,000, (b) 100,000, (c) 30,240, (d) 69,760, How many words can be formed with the letters, of the word MATHEMATICS by rearranging them, [MP PET 1984; DCE 2001], , 47., , 48., , 49., , 11 !, (a), 2!2!, , 11 !, (b), 2!, , 11 !, (c), 2!2!2!, , (d) 11 !, , [RPET 1990], , 52., , Circular permutations, 1., , 2., , (c), 51., , (2 ! ), , 9!, 2!, , (b), , (c) 9 !  2, , (d) None of these, , [Pb. CET 2000], , In how many ways can 5 keys be put in a ring, 1, 4!, 2, , (c) 4 !, 3., , 4., , 5., , 7., , (2 ! ) 3, , (d) 9 !, , (d) 5 !, , [, , (b) 4 !  5 !, (d) None of these, , (b) 10 !, , (c) 3 ! 10 !, (d) 3 ! 9 !, In how many ways can 15 members of a council, [MP PET 1984], sit along a circular table, when the Secretary is to, sit on one side of the Chairman and the Deputy, Secretary on the other side, (a) 2  12 !, (b) 24, (d) None of these, , In how many ways a garland can be made from, exactly 10 flowers, [MP PET 1984], (a) 10 !, (b) 9 !, (d), , 9!, 2, , 20 persons are invited for a party. In how many, different ways can they and the host be seated at, a circular table, if the two particular persons are, to be seated on either side of the host, (a) 20 !, (b) 2 . 18 !, (c) 18 !, , 8., , 1, 5!, 2, , In how many ways can 12 gentlemen sit around a, round table so that three specified gentlemen are, always together, , (c) 2(9 !), , 9!, , How many numbers can be made with the digits, 3, 4, 5, 6, 7, 8 lying between 3000 and 4000, , 5 ! 5 !, 2, , (c) 2  15 !, 6., , (b), , In how many ways can 5 boys and 5 girls sit in a, circle so that no two boys sit together, , (a) 9 !, , 2002], 2, , (b) 10 !, , (c), , (a) 66400, (b) 86400, (c) 96400, (d) None of these, How many words can be made from the letters of, the word COMMITTEE, [RPET 1986; MP PET, 9!, , (a) 10 !  2, , (a) 5! 5 !, , The number of arrangements of the letters of the, word CALCUTTA, [MP PET 1984], (a) 2520, (b) 5040, (c) 10,080, (d) 40,320, How many numbers, lying between 99 and 1000, be made from the digits 2, 3, 7, 0, 8, 6 when the, digits occur only once in each number, (a) 100, (b) 90, (c) 120, (d) 80, In a circus there are ten cages for accommodating, ten animals. Out of these four cages are so small, that five out of 10 animals cannot enter into them., In how many ways will it be possible to, accommodate ten animals in these ten cages, , (a), , If eleven members of a committee sit at a round, table so that the President and Secretary always, sit together, then the number of arrangements is, , (a), , [Roorkee 1989], , 50., , (a) 60, (b) 12, (c) 120, (d) 24, The letters of the word MODESTY are written in, all possible orders and these words are written, out as in a dictionary, then the rank of the word, MODESTY is, (a) 5040, (b) 720, (c) 1681, (d) 2520, , (d) None of these, , The number of ways in which 5 beads of different, colours form a necklace is, , [

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Permutations and Combinations, , 9., , PHOTON INSTITUTE UJJJAIN 8890244575/8502976721, , (a) 12, (b) 24, (c) 120, (d) 60, n gentlemen can be made to sit on a round table, in, , (a) 36, (b) 108, (c) 99, (d) 165, In how many ways can a girl and a boy be selected, from a group of 15 boys and 8 girls, (a) 15  8, (b) 15  8, , 4., , [MP PET 1982], , 1, (a) (n  1) ! ways, 2, , (b) (n  1) ! ways, , 1, (n  1) ! ways, 2, , (d) (n  1) ! ways, , (c), 10., , 5., , If, , The number of ways in which 5 male and 2 female, members of a committee can be seated around a, round table so that the two female are not seated, together is, (a) 480, (b) 600, (c) 720, (d) 840, In how many ways 7 men and 7 women can be, seated around a round table such that no two, women, can, sit, together, , (a) (7 !), , (c) (6 !) 2, 12., , 6., , 7., , 8., , (d) 7 !, permutations, , of, , 14., , (a) n!, (b) n, (c) (n – 2)!, (d) (n – 1)!, The number of ways that 8 beads of different, colours be string as a necklace is, (a) 2520, (b) 2880, (c) 5040, (d) 4320, The number of ways in which 6 men and 5 women, can dine at a round table if no two women are to, sit together is given by, (a) 6! × 5!, (b) 30, (c) 5! × 4!, (d) 7! × 5!, , If n is even and the value of, then r , n, (a), 2, n 1, (c), 2, , 2., , 3., , n, , C r is maximum,, , n 1, 2, , 5, , r 1, , 52, , (a), , 47, , C6, , (b), , (c), , 52, , C4, , (d) None of these, , n, , C5, , C r  C r 1 , n, , [MP PET 1984], , (a), , nr, r, , (b), , n  r 1, r, , (c), , n r 1, r, , (d), , n r 1, r, , 2n, , If, , C 3 : n C 2  44 : 3 ,, , then for which of the, n, , C r will be 15, , (b) r  4, (d) r  5, , (a) r  3, (c) r  6, 9., , If 2n C5  9 , , n2, , C5 , then the value of n will be, , (a) 7, (c) 9 [EAMCET 2002], 10., , If, , n 2 n, , C 2 n, , 2, , n, , 11., , (b) 10, (d) 5, , C10 , then n , , (a) 12, (c) 3 only, , (b) 4 only, (d) 4 or 3, , n, [AIEEE, RPET, 2003], If n C r2003;, 1  36 , C r  84 and, , value of r is, , n, , C r 1  126 , then the, , [IIT 1979; Pb. CET 1993, 2003;, , DCE 1999, 2000; MP PET 2001], , (a) 1, (c) 3, 12., , n, , (b) 2, (d) None of these, , C r  2 n C r 1  n C r  2 , , (c), , n 1, n2, , Cr, , (b), , Cr, , (d), , n 1, n2, , C r 1, C r 1, , 13., , In a conference of 8 persons, if each person shake, hand with the other one only, then the total, number of shake hands shall be, (a) 64, (b) 56, (c) 49, (d) 28, , 14., , n, , (d) None of these, , A man has 7 friends. In how many ways he can, invite one or more of them for a tea party, (a) 128, (b) 256, (c) 127, (d) 130, There are 12 volleyball players in all in a college,, out of which a team of 9 players is to be formed., If the captain always remains the same, then in, how many ways can the team be formed, , (b) 4, (d) 8, , C 4   52 r C 3  [IIT 1980; RPET 2002; UPSEAT 2000], , (a), (b), , C 3 r  C r  3 , then the value of r is, , [MP PET 1981], , Definition of combination, Conditional combinations,, Division into groups, Derangements, 1., , C2, , following values of r , the value of, , n, , [Kerala (Engg.) 2001], , 13., , 47, , (b) 7 ! 6 !, , The number of circular, different objects is, , 23, , 15, , (a) 3, (c) 5, , [EAMCET 1990; MP PET 2001;, DCE 2001; UPSEAT 2002;Pb. CET 2000], 2, , 15, , (d), , P2, , [IIT 1967; RPET 1991; MP PET 1998; Karnataka CET 1996], , [Roorkee 1999], , 11., , 23, , (c), , C r  n C r 1 is equal to, [MP PET 1984; Kerala (Engg.) 2002], , (a), (c), , n 1, n 1, , Cr, , (b), , C r 1, , (d), , n, , C r 1, , n 1, , C r 1

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Permutations and Combinations, 15., , If, , 8, , C r  8 C r  2 , then the value of r C 2 is, [MP PET 1984; RPET 1987], , (a) 8, (c) 5, 16., , If, , 20, , (b) 3, (d) 2, , C n  2  n C16 , then the value of n is [MP PET 1984], , (a) 7, (c) 13, 17., , 18., , (b) 10, (d) No value, , The value of, (a), , 16, , (c), , 15, , 15, , C 3  15 C 13 is, , C3, C 10, , (b), , 30, , C 16, , (d), , 15, , C 15, , (c) 8, d) None of these, How many triangles can be formed by joining four, points on a circle, (a) 4, (b) 6, (c) 8, (d) 10, 5., How many triangles can be drawn by means of 9, non-collinear points, (a) 84, (b) 72, (c) 144, (d) 126, [MP PET 1983], 6., The number of diagonals in a polygon of m sides, is, 4., , [BIT 1992; MP PET 1999; UPSEAT 1999;, DCE 1999; Pb. CET, , Everybody in a room shakes hand with everybody, else. The total number of hand shakes is 66. The, total number of persons in the room is, , 2001], , (a), , 1, m (m  5), 2!, , (b), , 1, m (m  1), 2!, , (c), , 1, m (m  3), 2!, , (d), , 1, m (m  2), 2!, , [MNR 1991; Kurukshetra CEE 1998; Kerala (Engg.) 2001], , (a) 11, (c) 13, 19., , (b) 12, (d) 14, , The solution set of, (a) {1, 2, 3}, (c) {8, 9, 10}, m, , 20., , , , n r, , 10, , C x 1  2 ., , 10, , C x is, , (b) {4, 5, 6}, (d) {9, 10, 11}, , Cn , , 8., , (a), (c), 21., , n m  3, , n m  2, , C n 1, , (b), , C n 1, , (d) None of these, , Cn, , In a football championship, there were played 153, matches. Every team played one match with each, other. The number of teams participating in the, championship is, , 22., , (a) 8, , (b) 16, , (c) 24, , (d) 28, , The number of triangles that can be formed by, choosing the vertices from a set of 12 points,, seven of which lie on the same straight line, is, [Roorkee 1989; BIT 1989; MP PET 1995;, , 9., , [WB JEE 1992; Kurukshetra CEE 1998], , (a) 17, (b) 18, (c) 9, (d) 13, In an examination there are three multiple choice, questions and each question has 4 choices., Number of ways in which a student can fail to get, all answers correct, is, , The number of straight lines joining 8 points on a, circle is, [MP PET 1984], , [Pb. CET 2003], , r 0, , n  m 1, , 7., , 10., , (a) 185, , (b) 175, , (c) 115, , (d) 105, , In a plane there are 10 points out of which 4 are, collinear, then the number of triangles that can be, formed by joining these points are, (a) 60, , (b) 116, , (c) 120, , (d) None of these, , There are 16 points in a plane out of which 6 are, collinear, then how many lines can be drawn by, joining these points, [RPET 1986; MP PET 1987], , [Pb. CET 1990; UPSEAT 2001], , (a) 11, (c) 27, , (b) 12, (d) 63, , Geometrical problems, 1., , 2., , 11., , The number of triangles that can be formed by 5, points in a line and 3 points on a parallel line is, 8, , C3 5 C3, , (a), , 8, , C3, , (b), , (c), , 8, , C3 5 C3  1, , (d) None of these, , The number of diagonals in a octagon will be, [MP PET 1984; Pb. CET 1989, 2000], , 3., , (a) 28, (b) 20, (c) 10, (d) 16, If a polygon has 44 diagonals, then the number of, its sides are, [MP PET 1998; Pb. CET 1996, 2002], (a) 7, (b) 11, , (a) 106, , (b) 105, , (c) 60, , (d) 55, , The straight lines I1 , I 2 , I 3 are parallel and lie in, the same plane. A total number of m points are, taken on I1 , n points on I 2 , k points on I 3 . The, maximum number of triangles formed, vertices at these points are, , with, , [IIT Screening 1993; UPSEAT 2001], , (a), (b), (c), , m n k, m n k, m, , C3, C 3 m C 3 n C 3  k C 3, , C3 nC3 k C3, , (d) None of these

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Permutations and Combinations, , 12., , 13., , PHOTON INSTITUTE UJJJAIN 8890244575/8502976721, , The number of parallelograms that can be formed, 11, c, from a set of four parallel lines intersecting, d, another set of three parallel lines is [WB JEE 1993; RPET162001], , 12, , c, , 13, , d, , 14, , a, , 15, , b, , 17, , a, , 18, , b, , 19, , c, , 20, , a, , (a) 6, , (b) 18, , 22, , d, , (c) 12, , (d) 9, , 21, , Six points in a plane be joined in all possible ways, by indefinite straight lines, and if no two of them, be coincident or parallel, and no three pass, through the same point (with the exception of the, original 6 points). The number of distinct points, of intersection is equal to, (a) 105, , (b) 45, , (c) 51, , (d) None of these, , Definition of permutation, Number of permutations, with or without repetition, Conditional permutations, 1, , b, , 2, , c, , 3, , a, , 4, , d, , 5, , b, , 6, , a, , 7, , b, , 8, , a, , 9, , c, , 10, , c, , 11, , d, , 12, , a, , 13, , d, , 14, , a, , 15, , a, , 16, , c, , 17, , c, , 18, , a, , 19, , d, , 20, , a, , 21, , c, , 22, , a, , 23, , c, , 24, , c, , 25, , a, , 26, , a, , 27, , d, , 28, , b, , 29, , b, , 30, , a, , 31, , c, , 32, , a, , 33, , c, , 34, , a, , 35, , b, , 36, , a, , 37, , b, , 38, , b, , 39, , c, , 40, , a, , 41, , a, , 42, , b, , 43, , b, , 44, , b, , 45, , d, , 46, , c, , 47, , b, , 48, , a, , 49, , b, , 50, , b, , 51, , b, , 52, , c, , 53, , b, , Circular permutations, , 1, , c, , 2, , a, , 3, , b, , 4, , d, , 5, , a, , 6, , d, , 7, , b, , 8, , a, , 9, , b, , 10, , a, , 11, , b, , 12, , d, , 13, , a, , 14, , a, , Definition of combinations, Condition combinations,, Division into groups, Derangements, 1, , a, , 2, , c, , 3, , d, , 4, , a, , 5, , a, , 6, , c, , 7, , c, , 8, , b, , 9, , b, , 10, , d, , b, , Geometrical problems, 1, , c, , 2, , b, , 3, , b, , 4, , a, , 5, , a, , 6, , c, , 7, , d, , 8, , a, , 9, , b, , 10, , a, , 11, , b, , 12, , b, , 13