3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.