8) The numbers 1, 2, 3, and 4 are drawn from an urn in any order. What is the probability that the first 3 numbers drawn are in increasing order?
(A) $\frac{1}{3}$
(B) $\frac{1}{4}$
(C) $\frac{1}{6}$
(D) $\frac{1}{8}$
(E) $\frac{1}{12}$.
8) The answer is $(\mathrm{C})$.

Regardless of the last number drawn, the probability that three numbers appear in ascending order is equal to the probability that the smallest among them is drawn first, which is $\frac{1}{3}$, multiplied by the probability that the smallest of the other two is drawn second, which is $\frac{1}{2}$.

Alternatively, one can observe that the total possibilities for the first three draws are $4 \cdot 3 \cdot 2=24$ (i.e., all possible ordered triples of numbers from 1 to 4), while the favorable cases are only 4 (i.e., the triples 123, 124, 134, 234). Therefore, the probability is $\frac{4}{24}=\frac{1}{6}$.