Question

If the probability of a six-digit number $N$ whose six digits are $1,2,3,4,5,6$ when written in random order, is divisible by $6$ is $p,$ then the value of $\frac{1}{p}$ is

Answer 1

Total number of cases $=n\left(S\right)=6!$
Now sum of the given digits is $1+2+3+4+5+6=21,$ which is divisible by $3.$
Now we have to form the number which is divisible by $6,$ then we have to ensure that the digit in unit place is even.
$\Rightarrow$ Favourable cases $=n\left(A\right)=3\cdot 5!$
Hence, $P\left(A\right)=\frac{3\cdot 5!}{6!}=\frac{1}{2}=p$
$\therefore \frac{1}{p}=2$