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Unit 13: Combinations




                                                                 r n – r
                                                                            b  will be the
          12.  In view of the fact that C(n, r) = C(n, n – r), the coefficients of a b   and a n – r  r  Notes
               ..............................,
          13.  The coefficient of   in the expansion (a  + a  +… + a ) is .............................., and is called a
                                                         n
                                              1  2      m
               multinomial coefficient, in analogy with the binomial coefficient.
          13.4 Inclusion-Exclusion Principle

          Let us begin with considering the following situation: In a sports club with 54 members, 34 play
          tennis, 22 play golf, and 10 play both. There are 11 members who play handball, of which 6 play
          tennis also, 4 play golf also, and 2 play both tennis and golf. How many play none of the three sports?
          To answer this, let S represent the set of all members of the club. Let T represent the set of tennis
          playing  members, G represent the set of golf playing members, and  H represent the set of
          handball playing members. Let us represent the number of elements in A by  A .


          Consider the number  S – T – G – H.
          Is this the answer to the  problem? No, because those  who are in  T as well as  G  have  been
          subtracted twice. To compensate for this double subtraction, we may now consider the number
           S   T   G   H   T   G   G   H   H  T .
          Is this the answer? No, because those playing all the three games have been subtracted thrice and then
          added thrice. But those members have to be totally excluded also. Hence, we now consider the number
                          S   T   G   H   T  G   G   H   H  T   T  G   H .

          This is the correct answer. This reduces to 54 – 34 – 22 – 11 + 10 + 6 + 4 – 2 = 5.
          To solve  this problem we have made inclusions and exclusions  alternately to arrive at  the
          correct answer. This is a simple case of the principle of inclusion and exclusion.



             Did u know?  It is also known as the sieve principle because we subject the objects to sieves
             of a progressively finer mesh to arrive at a certain grading.
          Let us state and prove this principle now.

          13.4.1 Inclusion-Exclusion Formula

          Let A , A ,…, A  be n sets in a universal set U consisting of N elements. Let S  denote the sum of
               1  2   n                                                k
          the sizes  of all the sets formed by intersecting k of the A is at a  time. Then  the number of
          elements in none of the sets A , A ,…, A  is given by
                                  1  2    n
                          A   A  ...  A n = N – S  + S  – S  +… +(–1)  S  +… +(–1) S n
                                                            k
                                                                       n
                          1
                              2
                                                    3
                                                2
                                                              k
                                             1
          Self Assessment
          Fill in the blanks:
          14.  Inclusion-Exclusion principle is also known as the .............................. principle because we
               subject the objects to sieves of a progressively finer mesh to arrive at a certain grading.
          15.  Inclusion-Exclusion formula is given by  A   A  ...  A n  = .............................. .
                                                 1
                                                     2

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