Number of ways to colour gridcells Given a $6\times 3$ grid, we would like to colour four cells with black in a way that exactly four of the columns contain one black square each.
How many ways are there to fulfill this colouring ?
My answer would be : $18 \times 15 \times 12 \times 9 $
18 possibilities to colour the first cell,
15 to colour the second,
12 to colour the third,
9 to colour the fouth.
Is my answer correct ?

Thanks
 A: Your answer is almost correct; you have overcounted by the number of different orders in which you can pick the same black squares. You can pick the same $4$ black squares in $4!$ different ways, so you should divide your current answer by $24$.
Another way to see this, is that you can pick $4$ different columns out of $6$ for the black squares, and then pick $1$ row out of $3$ for each of the $4$ black squares. This gives you $\binom{6}{4}\times3^4$.
A: No, your answer is not correct. Let's say that the first cells you color in order are $(1,1), (1,2), (1,3), (1,4)$. Then, in another attempt you can color them in another order, say in reverse, $(1,4), (1,3), (1,2), (1,1)$. See that these two approaches give you the same configuration. Therefore, by using your approach, you have counted some configurations twice.
The correct way to solve this problem is to first pick which columns are going to contain black cells. This can be done in $\binom{6}{4}$ ways. Then, you need to fill one cell of each of these rows, which can be done in $3^4$ ways. Applying the multiplicative rule, you get the result $\binom{6}{4}\cdot 3^4$.
