There is a grid of 3 columns: the first column has 4 squares, the second has 2 squares, and the third has 2 squares.
There are 6 marbles of different colors. How many ways can those 6 marbles be placed inside the grid such that each column has at least one marble?
My solution is as follows: There will be four cases as: $(4,1,1), (3,2,1), (3,1,2), (2,2,2)$ for number of marbles in respective columns. $$\bigg(\binom64 \cdot 4! \cdot \binom21 \cdot 2 \cdot 2\bigg) + 2 \cdot \bigg(\binom63 \cdot \binom43 \cdot 3! \cdot \binom32 \cdot 2 \cdot 2\bigg) \\ + \bigg(\binom62 \cdot \binom42 \cdot 2! \cdot \binom42 \cdot 2 \cdot 2\bigg) = 18720$$ (Logic: $\big(\binom62 \cdot \binom42 \cdot 2! \cdot \binom42 \cdot 2 \cdot 2\big)$ means first choose 2 balls out of 6, then choose 2 squares out of 4 in the first column and permute; second, choose 2 balls out of the 4 remaining balls and permute in the second column; in the third column, permute the 2 remaining balls).
Is this right or not? Kindly help me.