Question

A seven digit number is formed by using $0,1,2,3,4,8,9$ without repetition. Then the probability that it is divisible by $4$ is

• A. $\frac{53\times 4!}{5\times 6!}$
• B. $\frac{53}{180}$
• C. $\frac{53}{6\times 6!}$
• D. None of these

Answer 1

The correct option is B $\frac{53}{180}$

Two or more than two digit number will be divisible by 4 if its last two
digit are divisible by 4 so we need the number of the form 04, 08, 12,
24, 28, 32, 40, 48, 80, 84, 92. There are twelve cases out of which in 5
cases like 04, 08, 20, 40, 80, total numbers are $=5\times 5!$ and numbers in other 7 cases are $4\times 4!\times 7$
Number of favorable cases are $=5\times 5!+4\times 7\times 4!$ $=(25+28)4!=53\times 4!$
Total number of 7 digit number $=7!-6!=6.6!$
$\therefore$ Required probability
$=\frac{53\times 4!}{6\times 6\times 5\times 4!}=\frac{53}{180}$