Combinatorics Problem with Chess 8 black and 8 white pawns (totally 16 pawns) are placed at random on a chess board (64 squares with 8 rows and 8 columns). What is the probability that there is at least one row which has exactly 3 black pawns and 3 white pawns?
I thought that with this problem, the denominator would be the C(16,64)*C(8,16) for the numerator - I just do not know how to divide the cases.
 A: Let $A_i$ represent the event that row $i$ has exactly three black and three white pawns.
We are asked to count $|A_1\cup A_2\cup\dots\cup A_8|$
By inclusion-exclusion this is 
$$=|A_1|+\dots+|A_8|-|A_1\cap A_2|-|A_1\cap A_3|-\dots-|A_7\cap A_8|+|A_1\cap A_2\cap A_3|+\dots$$
By symmetry, we notice that $|A_1|=|A_2|=\dots=|A_8|$ and that $|A_1\cap A_2|=|A_1\cap A_3|=\dots=|A_7\cap A_8|$ etc... so the above simplifies to:
$$=8|A_1|-28|A_1\cap A_2|+|A_1\cap A_2\cap A_3|+\dots$$
Counting $|A_1|$:


*

*Arrange $3$ black and $3$ white pawns in the first row.

*Arrange the remaining $5$ black and $5$ white pawns in the remaining rows.

 This can be done in $\binom{8}{3}\binom{5}{3}\binom{56}{5}\binom{51}{5}$ ways

Counting $|A_1\cap A_2|$


*

*Arrange $3$ black and $3$ white pawns in the first row.

*Arrange $3$ black and $3$ white pawns in the second row.

*Arrange the remaining $2$ black and $2$ white pawns in the remaining rows.
Now, notice that it is impossible to have exactly three black and three white pawns in more than two rows at a time as that would require at least nine black pawns, which is more than our eight black pawns we have available.  This implies that all terms consisting of intersections of three or more of our events are empty and contribute zero to the overall sum.
Finish the arithmetic and take the appropriate ratio and reach a final answer.
