#
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.

$\left(a\right)1/25\phantom{\rule{0ex}{0ex}}\left(b\right)25/567\phantom{\rule{0ex}{0ex}}\left(c\right)1/37\phantom{\rule{0ex}{0ex}}\left(d\right)1/74$

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

**
**