Questions on Algebra: Complex Numbers answered by real tutors!

In order for ' f ' in the left side be a function of real argument,  must be defined.

It requires for ' x ' to be a non-negative real number  x >= 0.


So, the domain of the function in the left side is the set  { x >= 0 }.


Then the range is the set of all real numbers  { y >= 1 }.    <<<---=== ANSWER

The floor function rounds a real number x down to the nearest integer 
less than or equal to it. In other words, it returns the largest integer that is not greater than x.



        Notice that the given function is NOT defined at  x = 3.

        Due to this reason, the limit of the given function
        can not be determined directly.

        Let's consider this issue from different points of view.



When x approaches to 3 from the right side,    approaches to 1 from the right side.

Therefore,  has then the value of 1 identically in vicinity of x= 3  on its right side.
 
Hence, function    in the numerator is then identically 0 (zero) in vicinity of  x = 3  on its right side.

Therefore, the right limit of    as x ---> 3  from the right side can be naturally defined as  0 (zero).



On the contrary, when x approaches to 3 from the left side,    approaches to 1 from the left side.

Hence, function    is then identically -1  in vicinity of  x= 3  on its left side.

Therefore, the left limit of   as x ---> 3  does not exist (is positive infinity).



Thus, as x ---> 3, the limit of the function    in the common sense does not exist.



ANSWER.  The given function is not defined at x = 3.

         The right limit of the given function at  x --> 3  is  (or can be defined as)  0 (zero);  

         the left limit of the given function at  x --> 3  does not exist;

         the limit of the given function at  x --> 3  in common sense does not exist and can not be defined.

Place 4 green balls along the straight line.
Make gaps between these four green balls.

     You will have three gaps between the 4 green balls 
PLUS one gap on the left before these balls 
PLUS one gap on the right after these balls.


We will place 3 red balls and 2 white balls in these 3 + 2 = 5 gaps.
We MUST place some of these 3 + 2 = 5 balls in the gaps between the green balls
and we CAN place some of these 5 balls in the gaps before and/or after 4 green balls.


                It is the key idea to the problem' solution.


The rest is simply an implementation of this idea and detailed consideration/outlining/listing/counting 
of all possible cases.


    +----------------------------------------------------------------------------------------+
    |           Let the gaps between the 4 green balls are numbered 1, 2, 3;                 |
    |   the gap before green balls has number 0 and the gap after green balls has number 4.  |
    +----------------------------------------------------------------------------------------+



(1)  This is the case when all 3 red balls and 2 white balls are placed in 3 gaps between the 4 green balls.

     We have these sub-cases

               gap  #       1    2    3

        balls in the gaps   RWR  R    W,  
                            R    RWR  W,
                            R    W    RWR,   
        PLUS
                            RWR  W    R,  
                            W    RWR  R,
                            W    R    RWR,   
        PLUS
                            WRW  R    R,  
                            R    WRW  R,
                            R    R    WRW.   

     It gives  3 + 3 + 3 = 9 possible arrangements.


     We also have these other sub-cases

               gap #        1    2    3

        balls in the gaps   2    2    1,   where  '2'  is  (R,W)  or (W,R) independently at every '2' appearance.

                            2    1    2,

                            1    2    2.   It gives  = 12 possible arrangements.


    So, case (1) gives 9 + 12 = 21 different possible arrangements.



(2)  This is the case when all 3 red and 2 white balls are placed in 5 gaps, one ball in each gap.

     It gives   =  = 10 different possible arrangements for case (2).



(3)  This is the case when 2 balls of different colors are placed in the gap '0'; 
     three remaining balls of 5 = 3R + 2W balls are placed in gaps 1, 2, and 3.

     It gives these arrangements

               gap  #       0   1    2    3   4

        balls in the gaps  RW   R    R    W,
                           RW   R    W    R,
                           RW   W    R    R,      (3 arrangements)
        PLUS
                           WR   R    R    W,
                           WR   R    W    R,
                           WR   W    R    R.      (3 arrangements).


    It gives 3 + 3 = 6  different possible arrangements for case (3).



(4)  This is the case when 2 balls of different colors are placed in the gap '4'; 
     three remaining balls of 5 balls (3R + 2W) are placed in gaps 1, 2, and 3.

     This case is SYMMETRIC to case (3).  It gives 6 other arrangements, symmetric to case (3).



(5)  This is the case when 1 ball R or W is placed in the gap '0'; 
     four remaining balls of 5 = 3R + 2W balls are placed in gaps 1, 2, and 3, 
     so as two balls (R and W) are placed into the same gap.

     It gives these arrangements

               gap  #       0   1    2    3   4

        balls in the gaps   R   R    W    RW,
                            R   W    RW   R,
                            R   RW   R    W,    

                            R   W    R    RW,
                            R   R    RW   W,
                            R   RW   W    R,    

                            R   R    W    WR,
                            R   W    WR   R,
                            R   WR   R    W,    

                            R   W    R    WR,
                            R   R    WR   W,
                            R   WR   W    R      (3*4 = 12 arrangements)


        PLUS
                            W   R    R    RW,
                            W   R    RW   R,
                            W   RW   R    R,    

                            W   R    R    WR,
                            W   R    WR   R,
                            W   WR   R    R.     (3*2 = 6 arrangements)


    It gives 12 + 6 = 18  different possible arrangements for case (5).



(6)  This is the case when 1 ball R or W is placed in the gap '4'; 
     four remaining balls of 5 = 3R + 2W balls are placed in gaps 1, 2, and 3, 
     so as two balls (R and W) are placed into the same gap.

     This case is SYMMETRIC to case (5).  It gives 18 other arrangements, symmetric to case (5).



(7)  From cases (1), (2), (3), (4), (5) and (6),  we have, in all,  

         21 + 10 + 6 + 6 + 18 + 18 = 79 different possible arrangements.


ANSWER.  Doing this way, I counted 79 different possible arrangements.

As usual in such problems, we consider two specific cars as one block/unit.


So, now we have circular permutations of 4 objects.

The number of such circular permutations is (4-1)! = 3! = 1*2*3 = 6.


Also, there are 2 permutations inside the block, giving 2*6 = 12 different/distinguishable 
arrangements.


It is how everything works for circular permutations.


But in our case, the parking places are  .


Therefore, for 4 objects, we should multiply 12 by 4, because with numbered places,
we can start from any of 4 places from 1 to 4, which gives 4 times as many arrangements
as for the simple circular permutations case.


Thus the final answer is 4*(2*3!) = 4*12 = 48 different distinguishable arrangements.    ANSWER.


                +--------------------------------------------------+
                |   Another solution - another way of reasoning.   |
                +--------------------------------------------------+


For the two specific paired cars, we have 4 possible positions in the circular lot of 5 places.

We also have 2 possible permutations inside the pair.


For each such configurations of the two specific cars, we have (5-2)! = 3! = 6 possible permutations 
of the remaining 3 cars in 3 remaining positions.


The product 4*(2*3!) = 4*(2*6) =  4*12 = 48  gives the same answer 48.