If the probability of a six-digit number N whose six digits are 1, 2, 3, 4, 5, 6 written as random order is divisible by 6 it p, then the value of 1/p is ________.

Total number of cases = n(S) = 6!

Now sum 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. 

$\begin{array}{l}\Rightarrow \text{ Favorable cases }=\mathrm{n}\left(\mathrm{A}\right)=3\cdot 5!\\ \text{ Hence, }\mathrm{P}\left(\mathrm{A}\right)=\frac{3\cdot 5!}{6!}=\frac{1}{2}\end{array}$