SOLUTION: How many different ways can two balls be drawn from a bag containing five balls? Each ball is a different color, and the first ball is replaced before the second one is selected.

This problem has two major cases: 
    - one    when the first ball is replaced before the second ball is selected, and
    - second when the first ball is NOT replaced before the second ball is selected.


Each of the two major cases has two sub-cases: 
    - one    when the order of balls in final pairs is important, and 
    - second when the order of balls in final pairs is NOT important.


So, the problem has 2x2 = 4 different cases to consider and 4 different reasoning 
to get 4 different solutions and answers.



Case 1.  If the first ball is replaced before the second one is selected, then
          there are 5 choices for the first ball and 5 choices for the second ball.

          Sub-case 1.  If the order of the balls in final pairs is important, then 
                       the number of different ways (different final pairs) is 5*5 = 25.

                       The answer for this sub-case is 25.

                       It is the number of different entries/cells in 5x5-matrix.


          Sub-case 2.  If the order of the balls in final pairs is NOT important, then 

                       we count  = 10 different final pairs with different colors
                       and add to them 5 pairs of the same color. 

                       The answer for this sub-case is 10 + 5 = 15.

                       It is the number of over-diagonal cells in 5x5-matrix plus 5 diagonal entries.



Case 2.  If the first ball is NOT replaced before the second one is selected, then
          there are 5 choices for the first ball and 4 choices for the second ball.

          Sub-case 1.  If the order of the balls in final pairs is important, then 
                       the number of different ways (different final pairs) is 5*4 = 20.

                       The answer for this sub-case is 20.

                       It is the number of different entries/cells in 5x5-matrix
                       with its diagonal excluded.


          Sub-case 2.  If the order of the balls in final pairs is NOT important, then 
                       the number of different ways (different final pairs) is  = 20/2 = 10.

                       In this case, we count  = 10 different final pairs with different colors.

                       The answer for this sub-case is 10.

                       It is the number of over-diagonal cells in 5x5-matrix; 
                       the diagonal entries are excluded.