Combinatorics - placing stones on the chessboard In how many ways can we place 6 red, 6 green and 6 blue stones on a 5 × 5 chessboard so that some row or column is entirely covered by stones of the same color? 
 A: HINT: I suggest using an inclusion-exclusion argument. Suppose that the first row is covered with red stones. There are $20$ ways to place the remaining red stone, $\binom{19}6$ ways to place the green stones, and $\binom{13}6$ ways to place the red stones, so there are $20\binom{19}6\binom{13}6$ arrangements that have the first row completely covered with red stones. The same is true for each of the $10$ row and columns, so there are $200\binom{19}6\binom{13}6$ arrangements that have a row or column completely covered with red stones.
Clearly there are the same number of arrangements that have a row or column completely covered with green stones, and the same number again that have a row or column completely covered with blue stones. 
Now calculate similarly how many arrangements have a row or column completely covered with green stones and a row or column completely covered with green stones. Do the same for red and blue and for green and blue.
Finally, calculate similarly how many arrangements have a row or column completely covered with green stones, a row or column completely covered with red stones, and a row or column completely covered with blue stones. Then apply the usual inclusion-exclusion calculation to get the desired result.
