Key Concepts and Formulas

• Permutations: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Permutations with Restrictions: If we have $n$ objects and $k$ of them must be in a specific order relative to each other, we can treat those $k$ objects as indistinguishable initially. The number of arrangements is then $\frac{n!}{k!}$, where $n!$ is the total number of arrangements without the restriction, and we divide by $k!$ to account for the arrangements of the $k$ objects among themselves.

Step-by-Step Solution

Step 1: Identify the vowels and consonants.

The word GARDEN has 6 letters: G, A, R, D, E, N. The vowels are A and E, and the consonants are G, R, D, N.

Step 2: Consider the total number of arrangements without restrictions.

If there were no restrictions, the number of ways to arrange the 6 letters in GARDEN would be $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

Step 3: Apply the restriction that the vowels must be in alphabetical order.

The vowels A and E must be in alphabetical order (A before E). Without this restriction, the vowels could be arranged in $2! = 2 \times 1 = 2$ ways (AE or EA). Only one of these arrangements (AE) satisfies the alphabetical order condition. Therefore, we divide the total number of arrangements (without restrictions) by the number of ways the vowels can be arranged among themselves, which is $2!$.

Step 4: Calculate the number of arrangements with the given restriction.

The number of ways to arrange the letters in GARDEN with the vowels in alphabetical order is $\frac{6!}{2!} = \frac{720}{2} = 360$.

Common Mistakes & Tips

• Forgetting the factorial: Make sure to calculate the factorial correctly, especially for larger numbers.
• Misidentifying vowels/consonants: Double-check which letters are vowels and which are consonants. This is crucial for correctly applying the restriction.
• Applying the formula incorrectly: Understand the formula $\frac{n!}{k!}$ and when to use it. It's specifically for cases where a subset of letters must maintain a certain relative order.

Summary

We started by finding the total number of arrangements of the letters in GARDEN without any restrictions, which is $6! = 720$. Then, recognizing that the two vowels, A and E, must be in alphabetical order, we divided the total number of arrangements by $2!$ (the number of ways to arrange the two vowels) to account for the restriction. This gave us $\frac{720}{2} = 360$ possible arrangements.

Final Answer

The final answer is $\boxed{360}$, which corresponds to option (C).