Simple (not for me) combinatorics question There are four balls of different colors, and four boxes of colors, same as those of the balls.What are the number of ways, in which, the balls, one each in a box, could be placed, such that a ball does not go to a box of its own color. 
I name the balls A,B,C,D.
Correspondingly, I name the boxes A,B,C,D.
(i) Now starting with ball A, I could select one of the boxes B,C,D in 3 different ways.
(ii) Let, I put ball A in box B.
(iii) Moving onto ball B, I could select one of A,C,D in 3 different ways. 
(iv) Let, I put ball B in box C.
(v) As, no ball goes into corresponding box, ball C must be put in box D, ball D must be put in box A. Number of ways to do this is 1
So, total no. of ways=3 x 3 x 1 = 9
9 is the answer given in my book (not the process, this is an exercise problem, even  no hints given).
But now, consider the following case
(i) Starting with ball A, I could select one of the boxes B,C,D in 3 different ways.
(ii) Let, I put ball A in box B.
(iii) Moving onto ball C, I could select one of A,D in 2 different ways (as ball C cannot go to box C). 
(iv) Let, I put ball C in box D.
(v) Then I select ball B; it could be put in either of the boxes A,C in 2 different    ways. Let it be C.
(vi) Lastly, ball D could be put in box A, in only 1 way.
So, total no. of ways=3 x 2 x 2 x 1 = 12
Two different answers---something (I don't know what) should be wrong in the second solution.

Both methods are wrong. You should simply list out all the solutions (there are so few!) and see which of your steps break down or cannot be justified. The correct solution that generalizes to many boxes and many balls will require the inclusion-exclusion principle, but I think that is really beyond you now.
