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.


There are only two configurations where placing $6$ marbles in such a grid would not result in having a marble in every column, both involving having all four marbles in the left hand column. So your answer is simply

$${8 \choose 6}-2 = 28-2 = 26$$

for place locations, and then multiply in the permutations on the marbles in the chosen places

$$26\cdot 6! = 26\cdot 720 = 18720$$

which agrees with your method.


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