89) A five digit number is chosen at random. The probability that all the digits are distinct and digits at odd place are odd and digits at even places are even as.
$$\begin{gathered} \left(a\right) 1/25 \\ \left(b\right) 25/567 \\ \left(c\right) 1/37 \\ \left(d\right) 1/74 \end{gathered}$$

Hi, 
The 5 digit place will be 
_ _ _ _ _ 
Now odd places are 1st,3rd,5th can be filled by numbers 1,3,5,7,9,
and 2nd , 4th place can be filled by 2,4,6,8, 0 
Now in odd places can be filled in 5*4*3 ways = 60 ways 
and even places can be filled in 5*4ways = 20 ways 
total number of ways = 9*9*8*7*6
So required number of ways = 60*20/ 9*9*8*7*6
Regards