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Unit 6: Probability



            (b)  There are two ways in which 4 men and 4 women can be seated                      Notes

                 MWMWMWMWMW or WMWMWMWMWM
                     The required number of permutations = 2 .4! 4! = 1,152


                   Example 15: There are 3 different books of economics, 4 different books of commerce
            and 5 different books of statistics. In how many ways these can be arranged on a  shelf when

            (a)  all the books are arranged at random,
            (b)  books of each subject are arranged together,
            (c)  books of only statistics are arranged together, and
            (d)  books of statistics and books of other subjects are arranged together?

            Solution.
            (a)  The required number of permutations = 12!
            (b)  The economics books can be arranged in 3! ways, commerce books in 4! ways and statistics
                 book in 5! ways. Further, the three groups can be arranged in  3! ways.   The required
                 number of permutations = 3! 4! 5! 3! =1,03,680.
            (c)  Consider 5 books of statistics as one book. Then 8 books can be arranged in 8! ways and 5
                 books of statistics can be arranged among themselves in 5! ways.
                      The required number of permutations = 8! 5! = 48,38,400.
            (d)  There are two groups which can be arranged in 2! ways. The books of other subjects can be
                 arranged in 7! ways and books of statistics can be arranged in 5! ways. Thus, the required
                 number of ways = 2! 7! 5! = 12,09,600.

            6.2.3 Combination

            When  no  attention is  given to  the order  of arrangement  of  the  selected objects,  we get  a
                                                                                     n
            combination. We know that the number of permutations of n objects taking r at a time is  P .
                                                                                       r
            Since r objects can be arranged in r! ways, therefore, there are r! permutations corresponding to
            one combination. Thus, the number of combinations of n objects taking r at a time, denoted by
                                                        n
            n                          n           n    P       ! n
             C , can be obtained by dividing  P  by r!, i.e.,  C =  r  =  .
              r                          r           r
                                                         ! r  ( ! r n r-  )!
                                                  n
                             n
                                   n
            Note:  (a)  Since  C = C    ,  therefore,  C   is  also  equal  to  the  combinations  of  n
                               r     n- r            r
                        objects taking (n - r) at a time.
                   (b)  The total  number of  combinations of  n  distinct  objects  taking 1,  2,  ......  n
                                                 n
                                                              n
                                                                     n
                                            n
                        respectively, at a time is  C + C + ......  + C = 2 - 1 .
                                              1     2           n
                   Example 16:
            (a)  In how many ways two balls can be selected from 8 balls?
            (b)  In how many ways a group of 12 persons can be divided into two groups of 7 and 5 persons
                 respectively?







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