The seven letters of the word PICTURE are arranged at random in a row. What is the probability that no two of its three vowels end up next to each other?

PICTURE has 7 different letters — vowels I, U, E and consonants P, $C$, T, R. Lay the 4 consonants out first: 4! = 24 orders. They leave 5 gaps (before, between and after), and no two vowels are adjacent exactly when the three vowels sit in three DIFFERENT gaps: $C(5,3)=10$ choices of gaps, then 3! = 6 orders within them. Favourable = $24\times 10\times 6=1440$ out of 7! = 5040, giving 1440/5040 = 2/7. A shorter route: only which positions the vowels take matters, and $C(5,3)/C(7,3)=10/35=2/7$. Trap 5/7 is the opposite event, 'at least two vowels adjacent'. Trap 1/7 answers the other extreme, 'all three vowels together'. Trap 1/21 = 240/5040 forgets that the three vowels can be ordered among their chosen gaps in 3! ways.