$14 \cdot 3$ Randomly select 6 different integers from $\{1,2,3, \cdots, 10\}$. Among these selections, the probability that the second smallest number is 3 is
(A) $\frac{1}{60}$.
(B) $\frac{1}{6}$.
(C) $\frac{1}{3}$.
(D) $\frac{1}{2}$.
(E) None of the above.
(37th American High School Mathematics Examination, 1986)
[Solution] The number of ways to choose 6 numbers out of 10 is $C_{10}^{6}$,
and the number of ways where the second smallest number is 3, involves choosing one from 1,2 and then choosing 4 from $4,5,6$, $\cdots, 10$, which is $C_{2}^{1} C_{7}^{4}$.
$\therefore$ The required probability $p=\frac{C_{2}^{1} C_{7}^{4}}{C_{10}^{6}}=\frac{C_{2}^{1} C_{7}^{3}}{C_{10}^{4}}=\frac{1}{3}$. Therefore, the answer is $(C)$.