Chance that last digit in the product is $1,3,7 $ or $9$ 
If $4$ whole numbers are taken at random and multiplied 
  together ,what is the chance that last digit in the 
  product is $1,3,7 $ or $9$ ?

$a.)\ \dfrac{15}{653} \\
b.)\ \dfrac{12}{542} \\
c.)\ \color{green}{\dfrac{16}{625}} \\
d.)\ \dfrac{17}{625} $
I did $\dfrac{\dbinom{4}{4}}{\dbinom{10}{4}}=\dfrac{1}{210}$, but the answer given in book is $c.)$
I look for a short and simple way.
I have studied maths upto $12$th grade.

The last digits won't be evenly distributed.  If any number ends in $5$ or an even number then the product ends in a $5$ or an even number.  To end in a $1,3,7,$ or $9$ all numbers must end with $1,3,7,$ or $9$.  The probability of that is $(4/10)^4 = (2/5)^4 = 16/625$.
