In order for (n-1)! to be divisible by n, there must be a combination of factors within (n-1)! that is a factor of n. Since (n-1)! is a product of all of the integers less than n, the only solutions are when n is a prime number.

To illustrate, let's look at option A, n = 11.
$\frac{(11-1)!}{11} = \frac{10*9*8*7*6*5*4*3*2*1}{11}$
There are no common factors between 10! and 11. Thus, A is an answer.

Option B, n = 12.
$\frac{(12-1)!}{12} = \frac{11*10*9*8*7*6*5*4*3*2*1}{12}$
Since 6*2 = 12, we know that (12-1)! is divisible by 12. Thus, B is NOT an answer.

Ideally you would see the pattern by now, but to test a few more:

Option C, n = 13
$\frac{(13-1)!}{13} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{13}$
Since 13 is a prime, there's aren't any factors other than itself and 1. Thus, C is an answer.

Option D, n = 14.
$\frac{(14-1)!}{14} = \frac{13*12*11*10*9*8*7*6*5*4*3*2*1}{14}$
Since 7*2 = 14, we know that (12-1)! is divisible by 14. Thus, B is NOT an answer.