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

${ }^{2002}{C}_{1001}=$ 

$no of zeros in \left(2002\right)!=\left[\frac{2002}{5}\right]+\left[\frac{2002}{5^2}\right]+\left[\frac{2002}{5^3}\right]+\left[\frac{2002}{5^4}\right]+\left[\frac{2002}{5^5}\right]=$

no. of zeroes in (2002)!are $400+50+16+3=499$ 

no. of zeros in (1001)!=$\left[\frac{1001}{5}\right]+\left[\frac{1001}{5^2}\right]+\left[\frac{1001}{5^3}\right]+\left[\frac{1001}{5^4}\right]$
$(1001!{)}^2=2(200+40+8+1)=498$
Hence no of zeroes is =499-498=1