Question

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 is

• A. $\frac{1}{25}$
• B. $\frac{25}{567}$
• C. $\frac{1}{37}$
• D. $\frac{1}{74}$

Answer 1

The correct option is B $\frac{25}{567}$

Odd digits = 1, 3, 5, 7, 9
Even digits = 0, 2, 4, 6, 8
Since, odd digits at odd place and even digits at even place
Places of odd digits = 3
and places of even digits = 2
$\therefore$ Favorable ways =${}^5{P}_3{\times }^5{P}_2$
= 1200
Total ways $=9\times 9\times 8\times 7\times 6$
$\therefore$ Required probability =$\frac{1200}{9\times 9\times 8\times 7\times 6}=\frac{25}{567}$