What is the value of \( \frac{19!}{17!\cdot2!}\times\frac{2}{19} \)?

Correct answer: 18

Using \(19!=19\times18\times17!\) and \(2!=2\),
\[
\frac{19!}{17!\cdot2!}\times\frac{2}{19}
=\frac{19\times18\times17!}{17!\times2}\times\frac{2}{19}=18.
\]
Therefore, the correct value is 18. The factors 19, \(17!\), and 2 cancel out. Exam tip: In factorial expressions, expand the larger factorial only up to the smaller factorial before cancelling common factors.