Question 112
The expression $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Find the value of
$$
\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor .
$$
$$
\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor .
$$