Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is :

We need to find the probability that a randomly selected word from all arrangements of GARDEN does NOT have vowels in alphabetical order.

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

Total arrangements = $$6! = 720$$.

In any arrangement of the 6 letters, the two vowels A and E occupy 2 of the 6 positions. These two vowels can appear in only 2 orders: A before E (alphabetical) or E before A (non-alphabetical).

By symmetry, exactly half of all arrangements have A before E (vowels in alphabetical order), and exactly half have E before A.

The probability that vowels are NOT in alphabetical order (i.e., E appears before A) is:

$$P = \frac{1}{2}$$

The correct answer is Option (1): $$\frac{1}{2}$$.