<s>[INST] AIME I 2007 Problem $10$. Spot the overcounting? 
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.

My solution:
We can choose the $3$ squares from the first column in $\binom{6}{3}=20$ ways. I will fix $3$ squares from row $1,2,3$ which are in column $1$ and multiply them later by $20$. Also we will only find arrangements for the first $3$ columns since the last one is determined by them.
Case $1$: The second column has no common squares with the first column
There is $1$ way to choose the $3$ squares. Next, we can choose any $3$ squares from column $3$, therefore there are $1\cdot \binom{6}{3} =20$ ways to do this.
Case $2$: The second column has exactly $1$ square common with the first columm
There are $3$ ways to choose the common square and $3$ ways to choose $2$ squares from the remaining $3$. Next, there are 5 valid squares (since the row with common square has $2$ shadings) and we can choose $3$.Therefore there are $3\cdot 3\cdot \binom{5}{3}=90$ ways to do this.
Case $3$: The second column has $2$ common squares.
There are $3$ ways to choose the $2$ common squares, then there are $3$ ways to choose the $1$remaining square from the $3$ other squares. Next we choose $3$ squares from $4$ squares from the third column($2$ common squares hence their rows are filled). Therefore, there are $3\cdot 3 \cdot \binom{4}{3}=36$ ways to do this.
Case $4$: Second column has all $3$ squares common with the first column.
There is just $1$ way to choose the squares for the second column. Next there is just $1$ way for the third column since $3$ common squares have their rows filled. Therefore, there is just $1$ way to do this.
Adding up all the cases,
$$20+90+36+1=147$$
Finally multiplying this by $20$,
$$147×20=2940$$.
However the answer is $1860$ for the number of arrangements. I want to know where I overcounted.
Thanks
 [/INST] You took into account that some rows already have two shaded squares, but you didn’t take into account that some rows don’t have any shaded squares at all yet, and you need to shade them in both of the last two columns. So in case $2$ you can only choose $2$ of $4$ squares, with the third one fixed, and in case $3$ you can only choose $1$ of $2$ squares, with the other two fixed.
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