Question

A five digit number is chosen at random. Then probability that all the digits are distinct and digits at odd places are odd and digits at even places are even is

• A. $\frac{3}{65}$
• B. $\frac{1}{75}$
• C. $\frac{2}{65}$
• D. $\frac{8}{75}$

Answer 1

Answer: (B)

$\frac{1}{75}$

$Odddigits=1,3,5,7,9$
$Evendigits=0,2,4,6,8$
$Since$ $odd$ $digits$ $must$ $come$ $at$ $odd$ $places$ $and$ $even$ $digits$ $at$ $even$ $places,$
$Places$ $for$ $odd$ $digits=3$ $and$ $places$ $for$ $even$ $digits=2$
$\therefore$ $Favourable$ $number$ $of$ $ways$ $=$ $5\times 5\times 4\times 4\times 3=1200.$
$Total$ $number$ $of$ $five$ $digit$ $numbers$ $=$ $9\times 10\times 10\times 10\times 10=90000.$
$Thus$ $probability$ $=\frac{1200}{90000}=\frac{1}{75}$