Question

A five digit number is selected at at random. Then the probability that the digits in the odd places are odd and in the even places are even (no digit being repeated) is $\frac{a}{b}$ then $a+b$ is

Answer 1

The odd places can be filled up in $5\times 4\times 3$ ways and the even places in $5\times 4$ ways.
$\therefore$ the favourable number of ways $=5\times 4\times 3\times 5\times 4=5^2.2^4.3$
The number of five digit number $9\times {10}^4$
Hence the required probability $=\frac{5^2\times 2^4\times 3}{9\times {10}^4}=\frac{1}{75}$