22. From $\{1,2,3, \cdots, 10\}$, six different integers are randomly selected. 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.
$\frac{22}{C}$
22. To choose six different integers from $\{1,2, \cdots, 10\}$, there are $C_{10}^{\mathrm{B}}=210$ different ways. However, if the second smallest number is 3, then one number must be chosen from $\{1,2\}$, and four numbers must be chosen from $\{4,5, \cdots, 10\}$. This results in $C_{2}^{1} C_{7}^{4}=7 C$ ways. Therefore, the required probability is $\frac{70}{210}=\frac{1}{3}$.