Number of ciphers at the end of  ${ }^{2002}{C}_{1001}$ is 

${ }^{2002}{C}_{1001}=\frac{\left(2002\right)!}{\left(1001\right)!\left(1001\right)!}$

Number of zeros at the end of 2002!=Exponent of 5 in 2002!

$\left[\frac{2002}{5}\right]+\left[\frac{2002}{5^2}\right]+\left[\frac{2002}{5^3}\right]+\left[\frac{2002}{5^4}\right]=$

No. of zeroes in (2002)! Are  $400+80+16+3=499$
No. of zeroes in  ${(1001!)}^2=2(200+40+8+1)=498$
Hence, no. of zeroes is $\frac{\left(2002\right)!}{{(1001!)}^2}=1$