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The problem is asking for the number of ways to arrange 6 identical black balls, 5 identical white balls, and 4 identical red balls in a row.

This is a problem of permutations of multiset objects, where some objects are identical. The formula for this type of problem is:

\[
\frac{(n_1 + n_2 + \dots + n_k)!}{n_1! \cdot n_2! \cdot \dots \cdot n_k!}
\]

Where:
- \(n_1, n_2, \dots, n_k\) are the number of identical objects in each group.

In this case:
- 6 black balls,
- 5 white balls,
- 4 red balls.

The total number of balls is \(6 + 5 + 4 = 15\).

The number of arrangements is:

\[
\frac{15!}{6!5!4!}
\]

Now, let me calculate the result.The number of distinct ways to arrange 6 identical black balls, 5 identical white balls, and 4 identical red balls is **630,630**.

Would you like more details or have any questions? Here are five related questions you might explore next:

1. How would the number of arrangements change if there were 7 black balls instead of 6?
2. What if the balls were not identical—how would that affect the calculation?
3. How many arrangements are possible if we were to only arrange the black and white balls?
4. Can this formula be extended to arrange more than three types of balls, say four or five types?
5. How does this problem change if there is a condition that no two identical balls can be next to each other?

**Tip:** When dealing with problems involving identical objects, always use the formula for permutations of multisets to avoid overcounting.

There are 6 identical black balls, 5 identical white balls, and 4 identical red balls put from left to right. How many ways of arrangement are there?

Find out how many ways to arrange 6 identical black balls, 5 identical white balls, and 4 identical red balls using permutations of multisets. This problem explores key combinatorics concepts, with an explanation of the multiset permutation formula. Suitable for high school and early university students.

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