NSejs

I like to calculate without calculator, so i just divided by 5 and then divided the result by 5 again.
$\lfloor 2002:5\rfloor = 400; ~~~~~~~~~~~~~~~400:5=80; ~~~~~~~~~~~~~~80:5=16; ~~~~~~~~~~~~~~~\lfloor 16:5\rfloor=3$

$\lfloor 1001:5\rfloor = 200; ~~~~~~~~~~~~~~~200:5=40; ~~~~~~~~~~~~~~~40:5=8; ~~~~~~~~~~~~~~~\lfloor 8:5\rfloor=1$

$400-200-200 = 0; ~~~~~~~~~~80-40-40 = 0; ~~~~~~~~~~16-8-8 = 0; ~~~~~~~~~~3-1-1 = 1$

$\text{No. of trailing zeroes in } n! = \left \lfloor {\dfrac{n}{5}} \right \rfloor + \left \lfloor {\dfrac{n}{25}} \right \rfloor + \left \lfloor {\dfrac{n}{125}} \right \rfloor + \left \lfloor {\dfrac{n}{625}} \right \rfloor \cdots$

$\implies \text{No. of trailing zeroes in } 2002! =\left \lfloor {\dfrac{2002}{5}} \right \rfloor + \left \lfloor {\dfrac{2002}{25}} \right \rfloor + \left \lfloor {\dfrac{2002}{125}} \right \rfloor + \left \lfloor {\dfrac{2002}{625}} \right \rfloor =499$

$\implies \text{No. of trailing zeroes in } 1001! =\left \lfloor {\dfrac{1001}{5}} \right \rfloor + \left \lfloor {\dfrac{1001}{25}} \right \rfloor + \left \lfloor {\dfrac{1001}{125}} \right \rfloor +\left \lfloor {\dfrac{1001}{625}} \right \rfloor =249$

$\implies \text{No. of trailing zeroes in }(1001!)^2 =249\times 2 =498$

$\implies \text{No. of trailing zeroes in } \dfrac{2002!}{(1001!)^2} = 489-488= \boxed{1}$

Note: $\lfloor \cdot \rfloor$ represents the floor function

$\text{I used a calculator 😈😜. The answer was \boxed{1}}$