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:
Step-by-Step Solution
Key Concept: For any two distinguishable items in a random permutation, the probability one precedes the other is $\dfrac{1}{2}.$ Equivalently, $\dfrac{6!/2}{6!}=\dfrac{1}{2}.$
GARDEN has $6$ distinct letters; vowels A, E.
Total arrangements: $6!=720.$
Vowels in alphabetical order (A before E): exactly $\dfrac{6!}{2}=360.$ (Equivalently, choose $2$ positions for vowels: only one way to place A,E in order; remaining $4$ letters fill the others in $4!$ ways. So $\binom{6}{2}\cdot 4!=15\cdot 24=360.$)
NOT in order: $360.$ $P=\dfrac{360}{720}=\dfrac{1}{2}.$
Correct Answer: 1