Counting ball selections with color constraints from four boxes

Key takeaway: This is a constrained counting problem. The main challenge is enforcing the color condition box by box and then combining the results across all four boxes.

Break the problem into one box at a time

Each box contains 3 distinct red balls and 2 distinct blue balls. We want to choose 10 balls total from the 4 boxes, with at least one red and one blue chosen from each box.

For one box, count the ways to choose a valid subset:

• 2 balls: $1R,1B$ gives $\binom31\binom21=6$
• 3 balls: $2R,1B$ gives $\binom32\binom21=6$, and $1R,2B$ gives $\binom31\binom22=3$, so total $9$
• 4 balls: $2R,2B$ gives $\binom32\binom22=3$, and $3R,1B$ gives $\binom33\binom21=2$, so total $5$
• 5 balls: $3R,2B$ gives $\binom33\binom22=1$

So the one-box generating polynomial is

$6x^2+9x^3+5x^4+x^5.$

Use the total selection condition

We need the coefficient of $x^{10}$ in

$(6x^2+9x^3+5x^4+x^5)^4.$

Factor out $x^2$ from each box:

$=(x^2)^4(6+9x+5x^2+x^3)^4=x^8(6+9x+5x^2+x^3)^4.$

So we need the coefficient of $x^2$ in $(6+9x+5x^2+x^3)^4$.

There are two ways to make degree 2:

1. Choose one box to contribute degree 2 and the other three to contribute degree 0.
2. Choose two boxes to contribute degree 1 each and the other two to contribute degree 0.

Compute each case:

• One degree-2 box: $\binom41\cdot 5\cdot 6^3=4\cdot5\cdot216=4320$
• Two degree-1 boxes: $\binom42\cdot 9^2\cdot 6^2=6\cdot81\cdot36=17496$

Add them:

$4320+17496=21816.$

Final answer

The number of ways is

$\boxed{21816}.$

Why the counting works

The color condition is enforced inside each box first, which prevents invalid selections from being counted. After that, the problem becomes a product of independent choices, which is exactly the kind of situation generating functions handle well. The coefficient extraction step then isolates the total of 10 balls.

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Pitfalls the pros know 👇 A common mistake is to count only how many balls are chosen from each box and ignore the fact that the balls are distinct. Because the balls are distinct, each pattern like $2R,1B$ must be multiplied by the correct binomial coefficients. Another error is forgetting that the condition applies to every box, not just the overall set. If one box contributes only red or only blue, that selection is invalid even if the total of 10 balls looks correct. Students also sometimes miss the 4-box symmetry when counting the degree patterns, which leads to undercounting.

What if the problem changes? If the problem changed to "choose 10 balls so that from each box at least one red ball is chosen" and blue balls were unrestricted, the one-box polynomial would change because subsets with only blue balls would no longer be forbidden, but subsets with no red balls would be excluded. For example, if a box has 3 red and 2 blue balls, the allowed counts would be all nonempty subsets minus the all-blue subsets. The method would still use generating functions or casework, but the valid patterns and the final coefficient would be different.

Tags: generating functions, binomial coefficient, coefficient extraction