16) A friend of mine has written a secret four-digit number using the digits 1, 2, 3, and 4 exactly once. Knowing that no digit occupies the position corresponding to its value (i.e., the first digit is not 1, the second is not 2, and so on), what is the probability that I will guess the number correctly on the first try?
(A) $\frac{1}{24}$
(B) $\frac{1}{9}$
(C) $\frac{1}{8}$
(D) $\frac{1}{81}$
(E) $\frac{1}{6}$.
16) The answer is (B).

Indeed, the possible numbers my friend thought of are: $2143,2341,2413,3142,3412,3421,4123,4312,4321$.