anyone about the chess problem? i figured out the second one?
A chessboard has 64 squares arranged into 8 rows and 8 columns...
1. how many ways can eight identical pawns be placed on chessboard so that no two are in the same row or column?
2. how many ways can two white rooks and two black rooks be placed on a chessboard so that no two rooks of the opposite color are in the same row or column?
--ALSO--
A local dairy offers 16 flavors of ice cream...
1. how many ways are there to purchase eight scoops of ice cream? isn't that 16 choose 8?
2. how many ways are there to purchase eight scoops of ice cream of eight different flavors?
With regard to the chessboard:
1. The answer is 8! Do you see why?
2. This one is more complicated and it seems you need to consider several cases.
a) No two rooks are in the same row or column.
b) The white rooks are in the same row but the black rooks are not in the same row or column.
c) The white rooks are in the same column but the black rooks are not in the same row or column.
d) Same as b-c but with white and black interchanged.
e) The white rooks are in the same row and the black rooks are in the same row.
f) Same as e) but with columns instead of rows.
g) The white rooks are in the same row and the black rooks are in the same column.
h) Same as g) but with white and black interchanged.
Clearly many of these cases have the same number of possibilities (for example, b and c), so it's not as bad as it may look. However, there is still some trickiness in the analysis so you have to be careful.
I'll try to get you started with case a). Let's place the white rooks first. There are C(8,2) ways to choose the columns for the white rooks and C(8,2) ways to choose the rows. Given 2 rows and 2 columns we have 4 squares at the intersections, but the rooks must go at diagonally opposite corners (2 pairs of corners), so the white rooks can be placed in C(8,2) * C(8,2) * 2 ways. The black rooks can then be placed similarly, but once the white rooks are placed we only have 6 choices for the rows and columns of the black rooks, so the black rooks can be placed in C(6,2) * C(6,2) * 2 ways. Putting it all together, the number of possibilities for case a) is
C(8,2) * C(8,2) * 2 * C(6,2) * C(6,2) * 2.
I'll let you try the other cases.