What is the value of \(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}\)?

Correct answer: \(\frac{8}{3}\)

Since \(0!=1\), \(1!=1\), \(2!=2\), and \(3!=6\), the expression becomes \(1+1+\frac{1}{2}+\frac{1}{6}\). Using \(\frac{1}{2}=\frac{3}{6}\), the sum is \(2+\frac{3}{6}+\frac{1}{6}=2+\frac{4}{6}=\frac{8}{3}\). The distractor \(\frac{5}{2}\) may result from leaving out the \(\frac{1}{3!}\) term. Exam tip: always write \(0!=1\) explicitly in factorial calculations.