Probability Analysis of Word Arrangements: Random Examples and Calculations

Imagine you have the word ldquo;RANDOMrdquo; and you decide to play a fun game with its letters. The game involves arranging the word in a random manner and analyzing certain conditions regarding specific letters within the arrangement. In this article, we will delve into the mathematical logic and probability involved in one such condition.

Understanding the Problem

The question at hand is: ldquo;If the word ldquo;RANDOMrdquo; is arranged randomly, what is the probability that there exist only two letters between the letters ldquo;Ardquo; and ldquo;Ordquo;?rdquo; To solve this, letrsquo;s break down the problem into manageable parts and use combinatorial techniques to derive the answer.

Calculating the Total Number of Permutations

First, letrsquo;s calculate the total number of unique permutations of the word ldquo;RANDOM.rdquo; The word ldquo;RANDOMrdquo; consists of 6 distinct letters, and thus, the total number of permutations is given by:

$$text{Total permutations} 6! 720$$

Selecting and Arranging Other Letters

Next, we need to consider the specific arrangement where the letters ldquo;Ardquo; and ldquo;Ordquo; have exactly two letters between them. To do this, we can first select 2 letters from the remaining 4 letters (R, M, D, N). The number of ways to arrange these 2 letters is:

$$text{Ways to select and arrange 2 out of 4} frac{4!}{2!} 12$$

Now, these 2 letters (including ldquo;Ardquo; and ldquo;Ordquo;) can be considered as a single unit, with ldquo;Ardquo; and ldquo;Ordquo; themselves having 2 possible arrangements:

$$text{Ways to arrange A and O} 2! 2$$

Therefore, the total number of ways to arrange ldquo;Ardquo; and ldquo;Ordquo; with 2 other letters between them is:

$$text{Total ways} frac{4!}{2!2!} 6$$

Multiplying by the 2 possible arrangements of ldquo;Ardquo; and ldquo;Ordquo; itself, we get:

$$text{Total ways} 6 times 2 12$$

Calculating the Probability

Finally, to find the probability, we need to divide the number of favorable outcomes by the total number of permutations:

$$text{Probability} frac{48}{720} frac{1}{15} approx 0.0667$$

Conclusion

This problem is a great example of how combinatorial mathematics can be applied to real-world scenarios. By understanding permutations and the constraints, we can calculate the probability of specific arrangements. Such exercises not only sharpen our logical thinking but also provide valuable insights into the broader field of probability theory and combinatorics.