Arranging Balls with Specific Conditions

In this article, we explore the problem of arranging five white balls (W) and three red balls (R) in a row such that no two red balls are adjacent. Through careful analysis and application of combinatorial principles, we'll determine the total number of possible arrangements that meet this condition.

Problem Statement

The problem states that we need to place five identical white balls and three identical red balls in a row, with the condition that no two red balls touch each other. To solve this, we will break down the process into several steps.

Step-by-Step Solution

Step 1: Arrange the White Balls

Since the white balls are identical, there is only one possible arrangement for the white balls. We arrange them in a row:

W W W W W

Step 2: Identify Gaps for Red Balls

When the white balls are placed in a row, they create gaps where the red balls can be placed. Consider the arrangement:

_ W _ W _ W _ W _ W _

Here, each underscore represents a gap where a red ball can be placed. With five white balls, we observe that there are six gaps in total.

Step 3: Place Red Balls in Gaps

We need to select three out of these six gaps for placing the red balls. Since the red balls are identical, we need to calculate the number of ways to choose 3 gaps from 6 gaps. This can be determined using the combination formula:

Combination Formula Application

The combination formula to determine the number of ways to choose 3 gaps out of 6 is given by:

[binom{n}{r} frac{n!}{r!(n-r)!}]

Here, n is the total number of gaps, which is 6, and r is the number of red balls, which is 3. Substituting these values, we get:

[binom{6}{3} frac{6!}{3!(6-3)!} frac{6 times 5 times 4}{3 times 2 times 1} 20]

Therefore, there are 20 possible arrangements of the five white balls and three red balls such that no two red balls are adjacent.

Conclusion

The final answer to the problem is that there are 20 possible arrangements of the five white balls and three red balls with the condition that no two red balls are adjacent.

Additional Considerations

As a further exploration, we can consider cases where the red balls must be adjacent or form a connected trio. If the arrangement requires the three red balls to be together, the total number of arrangements changes:

First, treat the three red balls as a single unit (RRR). This reduces the problem to arranging one RRR and five Ws, which gives us 6 different arrangements. Additionally, there are 56 total arrangements considering both connected and disconnected red balls, as the total number of ways to arrange three indistinguishable red balls and five indistinguishable white balls is given by: [binom{8}{3} 56]

This exploration provides a more comprehensive understanding of the problem and its various interpretations.