We are asked to find the number of ways to arrange 6 identical black balls, 5 identical white balls, and 4 identical red balls from left to right.

Since the balls of the same color are identical, this is a combinatorics problem involving the arrangement of indistinguishable objects. We can solve this using the stars and bars method, which is a technique for distributing indistinguishable objects into distinguishable bins.

Total number of balls:

$6 \, (\text{black}) + 5 \, (\text{white}) + 4 \, (\text{red}) = 15 \, \text{balls}$

Now, we need to determine how to arrange 15 balls where 6 are black, 5 are white, and 4 are red. This is equivalent to finding the number of ways to arrange 15 objects where some of them are identical.

The formula for the number of ways to arrange $n$ objects where $a$, $b$, and $c$ objects are identical is given by: $\frac{n!}{a!b!c!}$ where $n$ is the total number of objects, and $a$, $b$, and $c$ are the counts of identical objects.

Here:

• $n = 15$
• $a = 6$ (black balls)
• $b = 5$ (white balls)
• $c = 4$ (red balls)

Substituting into the formula: $\frac{15!}{6!5!4!}$

We can now calculate this expression.

First, compute the factorials:

• $15! = 1,307,674,368,000$
• $6! = 720$
• $5! = 120$
• $4! = 24$

Now, substitute the values: $\frac{15!}{6!5!4!} = \frac{1,307,674,368,000}{720 \times 120 \times 24}$ $= \frac{1,307,674,368,000}{2,073,600} = 630$

Thus, the number of ways to arrange the balls is 630.

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Would you like further clarification on any step? Here are five related questions to consider:

1. How would the solution change if the balls were distinguishable instead of identical?
2. What if there were 3 different colors but an uneven distribution of balls?
3. How does the stars and bars method apply to problems with more than 3 categories?
4. Can you extend this method to problems involving non-ball objects, such as people or items?
5. How do factorials work, and why are they useful in combinatorics?

Tip: Always double-check your factorial calculations, as they grow rapidly and can affect the accuracy of combinatorial problems.