Answer
$15$
Work Step by Step
Since, we have $\sum_{i=1}^{5}[\dfrac{i!}{(i-1)!}]=\dfrac{1!}{0!}+\dfrac{2!}{1!}+\dfrac{3!}{2!}+\dfrac{4!}{3!}+\dfrac{5!}{4!}$
or, $=-1+\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{240}$
Hence, we get
$=1+2+3+4+5$
or, $=15$
