Tiêu đề XI - maths - chapter 7 - COMBIMATIONS(30-61).pdf ✅

30 NARAYANAGROUP PERMUTATIONS & COMBINATIONS JEE-MAIN SR.MATHS-VOL-II COMBINATIONS  A selection that can be formed by taking some or all of a finite set of things (or objects) is called a combination.  Formation of a combination by taking ‘r’ elements from a finite set A containing ‘n’ elements, means picking up r-elements subset of A n r  .  In case of combination, order of the objects is not important.  The number of combinations of ‘n’ dissimilar things taken ‘r’ at a time is equal to the number of ‘r’ elements subsets of a set containing ‘n’ elements.  The number of combinations of ‘n’ dissimilar things taken ‘r’ at a time is denoted by nCr or C (n, r) or C             n r n r or .  n Cr = !( )! ! r n r n  = r n n n n r r pr n 1.2.3... ( 1)( 2) ...( 1) !       0 1 1 , , n n n n C C C C   n n 2 2 , ......, n n n n C C C C   n r n r   W.E-1: If there are 12 persons in a party, and if each two of them shake hands with each other, the number of handshakes in the party is Sol : It is to note that, when two persons shake hands, it is counted as one handshake. The total number of handshakes is same as the number of ways of selecting 2 persons among 12 persons. 12 2 12! 66 10! 2!    C   0 1 2 ......... 2 n n n n n C C C C      n        0 1 2 2 1 2 1 ......... 2 1 2 n n C C c n n n                       1 2 2 1 2 2 1 2 1 ......... 2 1 2 n n n n C C c n n n  n Cr + n Cr-1 = (n+1)Cr  If n Cr = n Cs then r = s or r+s = n.  ( 1) ( 1) , 2 2 1 1         r r n n C C r n C C r n r n r n r n , etc.  r n r C C r n r n 1 1      n r n C C r n r n   ( 1) W.E-2: The least value of ‘ n ’ so that  1 1    3 4 3 n n n C C C     is [Eamcet 2011] Sol.  1 1    3 4 3 n n n C C C     4 3 4 3 n n      C C n  The least value of n is 8. W.E-3:If 1 1 330, 462, 462 n n n C C C r r r r        (Eamcet-2013) Sol. 1 462 1 7 330 5 n r n r C n r C r              5 5 5 7 5 12 5 n r r n r 1 462 1 462 1 n r n r C n r C r       n r r n r        1 2 1 5 12 5 5 2 1 12 5 n r r r                 10 5 12 5 2 10 5 r r r r  If n Cr is greatest, then r = 2 n when n is even, r = 2 1 2 1 n  or n when n is odd  Vandermonde’s Theorem : mC0 . n Cr + mC1 . n Cr-1 + mC2 . n Cr-2 + .... ..... + mCr-1 . n C1 + mCr . n C0 = m+nCr . W.E-4: The sum 0 10 20 m r i m i                is maximum when m is Sol: 10 20 10 20 0 0 M m i m i I i i m i C C                    10 10 10 10 20 20 20 20 0 1 1 0 .... C C C C C C C m m m m       30Cm is maximum when 30 15 2 m   SYNOPSIS
NARAYANAGROUP 31 JEE-MAIN SR.MATHS-VOL-II PERMUTATIONS & COMBINATIONS   0,  n C r n r  is divisible by ‘ n ’ only if ‘ n ’ is prime number. As 6 C2 is not divisible by 6 but 7 C4 is divisible by 7.  The number of combinations of ‘n’ things taken ‘r’ at a time in which a) ‘s’ particular things will always occur is (n-s)Cr-s b) ‘s’ particular things will never occur is n-sCr c) ‘s’ particular things always occur and ‘p’ particular things never occur is (n-p-s)Cr-s.  If ‘n’ different objects are in a row then the number of ways of selecting ‘r’ objects at a time so that no two of these ‘r’ objects are consecutive is  1Cr n r    If ‘n’ different objects are on a circle then the number of ways of selecting ‘r’ objects at a time so that no two of these ‘ r’ objects are consecutive is     2 1 1 r r C C n r n r       W.E.5: In how many ways can a cricket eleven be chosen out of a batch of 15 players if (a) There is no restriction on the selection (b) a particular player is always chosen (c) a particular player is never chosen ? Sol :(i) The total number of ways of selecting 11 players out of 15 is 15 15 15 11 15 11 4 15 14 13 12 1365 4 3 2 1 C C C             (ii) if a particular player is always chosen. This means that 10 players are selected out of the remaining 14 players. Hence Required number of ways 14 14 14     C C C 10 14 10 4  1001 (iii) If a particular player is never chosen. This means that 11 players are selected out of the remaining 14 players. Hence Required number of ways 14 14 14     C C C 11 14 11 3  364 W.E.6: The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is Sol : Required number of ways (n-p-s) 22-4-2 16 C = C r-s 11-2 9  C Groups:  Number of ways in which (m+n) items can be divided into two unequal groups containing ‘m’ and ‘n’ items is m+nCm = ! ! ( )! m n m  n  Number of ways of distributing (m + n) different things to two persons so that one gets ‘m’ things and the other gets ‘n’ things is  ! 2! ! ! m n m n    Number of ways of dividing 2n different things into two groups, each containing ‘ n ’ things and the order of the groups is not important, is     2 2 2! ! n n .  Number of ways of dividing, 2n different things in two groups, each containing ‘ n ’things and the order of the groups is important, is     2 2 ! ! n n .  The number of ways in which ‘mn’ different items can be divided equally into ‘m’ groups, each containing ‘n’ objects and the order of the groups is not important is ( )! 1 ( !) ! m mn n m        The number of ways in which ‘mn’ different objects can be divided equally into ‘m’ groups, each containing ‘n’ objects and the order of groups is important, is ( )! 1 ( !) ! m mn n m               m! = ( )! ( !)m mn n  The number of ways in which (m + n + p) things can be divided into three different groups of m, n and p things respectively is !. !. ! ( )! m n p m  n  p  The required number of ways of dividing ‘3n’ things into three groups of ‘n’ each = 3! 1 . !. !. ! (3 )! n n n n When the order of groups has importance then required number of ways = (3n)! / (n!)3 .  The number of ways in which ‘n’ different things can be arranged into ‘r’ different groups is n+r-1Pn or n!. n-1Cr-1 according to blank groups are admissible or not admissible.
32 NARAYANAGROUP PERMUTATIONS & COMBINATIONS JEE-MAIN SR.MATHS-VOL-II  The number of ways in which ‘n’ different things can be distributed into ‘r’ different groups is rn - r C1 (r-1)n + r C2 (r-2)n - ... + (-1)r-1 1 r Cr Here blank groups are not allowed.  Distribution of ‘ n ’distinct objects in ‘r ’different boxes if in any box any number of objects are placed (empty boxes are allowed) Here, each ojbect has ‘r ’ possibilities. Hence, number of ways .... n      r r r r n times r . W.E-7: The number of ways can a pack of 52 cards be divided equally among four players in order is Sol : For the first player we have 52C13 choices, for the second player 39C13 choices, for the third player 26C13 choices and for the last player we have 13C13 choices. Hence, the total number of ways   52 39 26 13 13 13 13 13 4 52! 13!      C C C C W.E-8: The number of ways of dividing 20 persons into 10 couples is Sol : Here the order of the couples is not important. So, required number of ways is   10 20! 2! 10!  If ‘n’ points are on the circumference of a cicle are given, then i) number of straight lines = C2 n . ii) number of triangles = C3 n iii) number of quadrilaterals = C4 n .... so on.  If a polygon has 'n' sides then the number of diagonals in it is n C2 - n (or) 2 n(n  3) .  In a plane there are 'n' points and no three of which are collinear except 'k' points which lie on a line. Then i) Number of straight lines that can be formed by joining them = n C2 - k C2 + 1. ii) Number of triangles that can be formed by joining them = n C3 - k C3 .  If a set of 'm' parallel lines are intersected by another set of 'n' parallel lines then the number of parallelograms that can be formed = (mC2 ).(n C2 )  Number of rectangles of any size in a square of nxn is  n r r 1 3 and number of squares of any size is  n r r 1 2 .  In a rectangle of n p  (n < p) number of rectangles of any size is 4 np (n+1) (p+1) and number of squares of any size is  n r 1 (n+1-r) (p+1-r).  Number of rectangles on a chess board (including squares) = 1296  Number of squares of all dimensions on a chess board = 204  Number of rectangles on a chess board which are not squares = 1092. W.E-9: Out of thirty points in a plane, eight of them are collinear. The number of straight lines that can be formed by joining these points is (Eamcet - 2014) Sol.The number of straight lines formed 30 8    C C 2 2 1= 408 W.E-10: The straight lines 1 2 3 l l l , , are parallel and lie in the same plane. A total number of m points are taken on 1 l n, points on 2l k, points on 3l . The maximum number of triangles formed with vertices of these points are Sol. Total number of points    m n k  Number of selections taking 3 at a time 3 m n k C    The points on the same line will not given any triangle.  Number of triangles 3 3 3 3 m n k m n k C C C C      
NARAYANAGROUP 33 JEE-MAIN SR.MATHS-VOL-II PERMUTATIONS & COMBINATIONS Problems based on certain theorems on combinations:  The total number of combinations of (p1 + p2 + .... +pk ) things taken any number at a time when ‘p1 ’ things are alike of one kind, p2 things are alike of second kind, ... pk things are alike of th k kind, is (p1 + 1) (p2 +1) .... (pk + 1).  The total number of combinations of (p1 +p2 +...+pk ) things taken one or more at a time when p1 things are alike of one kind, p2 things are alike of second kind, .... pk things are alike of th k kind, is (p1 + 1) (p2 + 1) ... (pk + 1) - 1.  The total number of combinations of ‘n’ different things taken any number at a time is 2n (or)  The number of ways of answering all of ‘n’ questions when each question has an alternative is 2n .  The total number of combinations of ‘n’ different things taken one or more at a time is 2n - 1. (or) The number of ways of answering one or more of ‘n’ questions is 2n - 1.  The number of ways of answering one or more of ‘n’ questions when each question has an alternative is 3n - 1. W.E-11: There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is Sol. Required number of ways 10 10 10 10 10 1 2 3 10        C C C C ........ 2 1 W.E-12: The number of different sums that can be formed with the coins a rupee, a 50 paise, a 25 paise, a 10 paise, and a 5 paise is Sol. One rupee coin can be selected in two ways. That is either selecting or non selecting a coin. Similarly the remaining coins also can be selected in two ways. In this process selecting none of the coins is selected is one way. The number of different sums is (1+1) (1+1) (1+1) (1+1) (1+1) -1 = 31 W.E-13: The number of ways in which ‘ n ’distinct objects can be put into two different boxes is (Each box can hold ‘n’ objects) Sol : Let the two boxes be B1 and B2 . There are two choices for each of the n objects. So, the total number of ways is2 2 .... 2n     n times W.E-14: The number of ways in which ‘ n ’ distinct objects can be put into two different boxes so that no box remains empty, is Sol : Each object can be put either in box B1(say) or in box B2 (say). So, there are two choices for each of the ‘n’ objects. Therefore the number of choices for ‘ n ’ distinct objects is 2 2 ......   upto n times 2n  . This includes either the first or the second box being empty. Thus 2 2 n  ways. W.E-15: The number of ways in which ‘ n ’ distinct objects can be put into two identical boxes so that no box remains empty, is Sol : We know that there are 2 2 n  ways in which neither box is empty. Since two boxes are identical, required no. of ways   1 2 2 2 n    Exponent of ‘P’ (P is prime) in n n N !   (OR) Highest power of P in n! is 2 3 ..... n n n P P P                      Where  x denotes greatest integer function  x W.E-16: Exponent of 3 in 100! is Sol : 2 3 4 5 100 100 100 100 100 ... 3 3 3 3 3                                           33 11 3 1 0 ...... 48  If n! 2 3 5 7 ....      where      ......... then n! ends with  zeros (i.e. that is exponent of 5). W.E.17:- No. of zeroes in 100! (or) Exponent of 5 in 100! Sol: 2 3 100 00 100 ....... 20 4 0 24 5 5 5                             Let N = k p p p pk     . . ...... 1 2 3 1 2 3 where p1 , p2 , p3 , ... pk are different primes and 1 ,  2 ,  3 ,...  k are natural numbers then  The total number of divisors of N = ( 1 + 1) ( 2 + 1) .... ( k + 1)  The total number of proper divisors of N (excluding 1 and N ) = ( 1 + 1) ( 2 + 1) .... ( k + 1) - 2