Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?
\(\frac{1}{24}\)
\(\frac{1}{6}\)
\(\frac{2}{13}\)
\(\frac{1}{4}\)
Correct answer is A
The number of arrangements for the 4 letters = \(^{4}P_{4} = \frac{4!}{(4 - 4)!}\)
\(4! = 24\)
Alphabetical order is just 1 of the arrangements for the letters
= \(\frac{1}{24}\)