We are tasked with finding the number of ways to order the letters of the alphabet such that no vowels are adjacent to each other.

Step 1: Identify vowels and consonants

In the English alphabet:

• Vowels: A, E, I, O, U (5 vowels)
• Consonants: The remaining letters are consonants. There are $26 - 5 = 21$ consonants.

Step 2: Arrange the consonants

First, we arrange the consonants. Since there are 21 consonants, they can be arranged in $21!$ ways.

Step 3: Place the vowels

We need to place the vowels in such a way that no two vowels are adjacent. To ensure this, we must place the vowels in the gaps between the consonants.

When the 21 consonants are arranged, there are 22 positions available to place the vowels: one before the first consonant, one between each pair of consonants, and one after the last consonant.

Step 4: Select positions for the vowels

Out of these 22 available positions, we need to select 5 positions to place the 5 vowels. The number of ways to select 5 positions from 22 is given by $\binom{22}{5}$.

Step 5: Arrange the vowels

After choosing 5 positions, the 5 vowels can be arranged in those positions in $5!$ ways.

Step 6: Calculate the total number of arrangements

The total number of ways to arrange the letters such that no two vowels are adjacent is: $21! \times \binom{22}{5} \times 5!$ Substituting the values: $21! \times \frac{22!}{5!(22-5)!} \times 5!$

Let's compute this.The total number of ways to arrange the letters of the alphabet such that no two vowels are adjacent is 161,451,464,537,975,567,155,200,000.

Would you like further clarification or a breakdown of any specific part?

Here are some related questions you might explore:

1. How many ways can the letters of the alphabet be arranged without any restriction?
2. What if the condition is no two consonants can be adjacent?
3. How many ways can we arrange only the vowels such that no two are adjacent in the alphabet?
4. How would the solution change if "Y" were also considered a vowel?
5. How many arrangements are there if only 10 consonants and 3 vowels are used?

Tip: Binomial coefficients, like $\binom{n}{r}$, are commonly used to count combinations when selecting positions or groups, a useful tool in combinatorics.