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Quantitative Techniques – I




                    Notes          3.  the total number of  ways of  simultaneous throwing of 2  coins and  a die is equal  to
                                       2   2   6 = 24.


                                          Example:
                                   1.  In how many ways can the letters of the word EDUCATION be arranged?
                                   2.  In how many ways can the letters of the word STATISTICS be arranged?

                                   3.  In how many ways can 20 students be allotted to 4 tutorial groups of 4, 5, 5 and 6 students
                                       respectively?
                                   4.  In how many ways 10 members of a committee can be seated at a round table if (a) they can
                                       sit anywhere (b) president and secretary must not sit next to each other?
                                   Solution:
                                   1.  The given word EDUCATION has 9 letters. Therefore, number of permutations of 9 letters
                                       is 9! = 3,62,880.
                                   2.  The word STATISTICS has 10 letters in which there are 3S’s, 3T’s, 2I’s, 1A and 1C. Thus, the
                                       required number of permutations = 50,400.
                                   3.  Required number of permutations = 9,77,72,87,522
                                   4.  (a)  Number of permutations when they can sit anywhere = (10-1)!= 9! = 3,62,880.
                                       (b)  We first find the number of permutations when president and secretary must sit
                                            together. For this we  consider president and secretary  as one person. Thus, the
                                            number of permutations of 9 persons at round table = 8! = 40,320.

                                      The number of permutations when president and secretary must not sit together = 3,62,880 –
                                   40,320 = 3,22,560.

                                   12.3.3 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 P    n!
                                   n                          n           n C   r
                                    C , can be obtained by dividing  P  by r!, i.e.,   r
                                     r                          r              r!  r! n r !
                                   Note:

                                             n
                                                               n
                                   1.  Since  C  n C  , therefore,  C   is also  equal to the  combinations of  n  objects taking
                                               r   n r           r
                                       (n - r) at a time.
                                   2.  The total number of combinations of n distinct objects taking 1, 2, ...... n respectively, at a
                                              n
                                       time is  C  n C   ......   n C  2 n  1 .
                                                1   2         n












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