Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are

Q: Why can't we just use the stars and bars formula for $n_1 + n_2 + n_3 + n_4 = 10$?

Stars and bars is excellent for distributing identical items into distinct bins. However, here the items (balls) are distinct, and the number of ways to pick them depends on the *number* of balls picked from each box. We must first determine the valid distributions of the *counts* of balls, and then calculate the combinations for each count separately.

Q: How do I know I found all the possible distributions?

We are partitioning $10$ into $4$ parts, with each part $\ge 2$. Let $n_i = 2 + x_i$, where $x_i \ge 0$. Then $(2+x_1) + (2+x_2) + (2+x_3) + (2+x_4) = 10$, which simplifies to $x_1 + x_2 + x_3 + x_4 = 2$. The possible non-negative integer solutions for $x_i$ are $(2,0,0,0)$ (which gives the distribution $4,2,2,2$) and $(1,1,0,0)$ (which gives $3,3,2,2$). These are the only two cases.

Q: Why do we multiply by the permutations of the boxes?

The boxes are distinct. If we have a distribution like $(4,2,2,2)$, it matters *which* box gets the $4$ balls. There are $4$ different boxes that could receive the $4$ balls, so we must multiply by the number of ways to arrange these counts among the boxes.

Question

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

Options:

(A) 21816

(B) 85536

(C) 12096

(D) 156816

Sigma Percentile · Accepted Answer

### The Combinatorial Journey: Unlocking the Mystery of the Boxes

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey of logical deduction.

We have four distinct boxes, each holding a treasure of $3$ red and $2$ blue balls. Our goal is to select $10$ balls such that every single box contributes at least one red and one blue ball.

This is not merely a counting exercise; it is a test of how we structure our reality.

### Phase 1: The Constraint of the Minimum

First, let us visualize the constraint. If every box must yield at least one red and one blue ball, then every box must contribute at least $2$ balls.

If we let $n_i$ be the number of balls chosen from box $i$, we are bound by the equation:

$$n_1 + n_2 + n_3 + n_4 = 10, 	ext{ where } n_i \ge 2$$

This is our anchor. We are partitioning the integer $10$ into $4$ parts, with each part being at least $2$.

This restriction is powerful; it collapses the infinite possibilities into just two distinct scenarios: the distribution $(4, 2, 2, 2)$ and the distribution $(3, 3, 2, 2)$. Any other combination, like $(5, 2, 2, 1)$, would violate our rule of $n_i \ge 2$.

### Phase 2: The Art of Selection

Now, we must determine how many ways we can pick $k$ balls from a single box, ensuring the condition of at least one red and one blue.

For $k=2$, we must pick $1$ red and $1$ blue. The number of ways is:

$$\binom{3}{1} 	imes \binom{2}{1} = 3 	imes 2 = 6$$

For $k=3$, we have two mutually exclusive paths: picking $2$ red and $1$ blue, or $1$ red and $2$ blue. The calculation is:

$$\binom{3}{2}\binom{2}{1} + \binom{3}{1}\binom{2}{2} = (3 	imes 2) + (3 	imes 1) = 9$$

For $k=4$, we can pick $3$ red and $1$ blue, or $2$ red and $2$ blue. This gives us:

$$\binom{3}{3}\binom{2}{1} + \binom{3}{2}\binom{2}{2} = (1 	imes 2) + (3 	imes 1) = 5$$

### Phase 3: The Permutation of Possibilities

We now have the building blocks. For Case 1, the distribution $(4, 2, 2, 2)$, we must account for the fact that the boxes are distinct.

The number of ways to assign these counts to the boxes is:

$$\frac{4!}{3!} = 4$$

For each assignment, we have $5$ ways to pick from the box with $4$ balls, and $6$ ways for each of the three boxes with $2$ balls. Thus, the total for Case 1 is:

$$4 	imes (5 	imes 6^3) = 4320$$

For Case 2, the distribution $(3, 3, 2, 2)$, the number of ways to assign these counts is:

$$\frac{4!}{2!2!} = 6$$

For each assignment, we have $9$ ways for each of the two boxes with $3$ balls, and $6$ ways for each of the two boxes with $2$ balls. The total for Case 2 is:

$$6 	imes (9^2 	imes 6^2) = 6 	imes 81 	imes 36 = 17496$$

### The Final Synthesis

We have reached the summit. Since these two cases are mutually exclusive, we simply add them together:

$$4320 + 17496 = 21816$$

The elegance of this result lies not in the number itself, but in the path we took to find it. We respected the constraints, partitioned the possibilities, and accounted for the distinct nature of our boxes.

The final answer is **21816**.

Select Answer:

Visualized Solution

Understanding the Setup

Minimum Balls per Box

Possible Distributions

Ways to Choose $2$ Balls

Ways to Choose $3$ Balls

Ways to Choose $4$ Balls

Analyzing Case 1: $(4, 2, 2, 2)$

Analyzing Case 2: $(3, 3, 2, 2)$

Final Calculation

The Sigma Insight: Combinations and Selection

The Combinatorial Journey

Phase 1

Phase 2

Phase 3

The Final Synthesis

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