$14 \cdot 6$ If $a, b, c$ are any two elements taken from the set $\{1,2,3,4,5\}$ (not necessarily distinct). Then the probability that $ab+c$ is even is
(A) $\frac{2}{5}$.
(B) $\frac{59}{125}$.
(C) $\frac{1}{2}$.
(D) $\frac{64}{125}$.
(E) $\frac{3}{5}$.
(46th American High School Mathematics Examination, 1995)
[Solution] First, the total number of ways to pick three elements from the set is $5^{3}=125$.
Now consider the case of $c$: There are $C_{3}^{1}=3$ ways to choose one from $\{1,3,5\}$, making $c$ an odd number; there are $C_{2}^{1}=2$ ways to choose one from $\{2,4\}$, making $c$ an even number.
The number of cases where $a b$ is odd is $3^{2}=9$, and the number of cases where $a b$ is even is $5^{2}-3^{2}=16$. By the rules “odd + odd = even” and “even + even = even”, the number of cases where $a b+c$ is even is
$$
3 \times 9 + 2 \times 16 = 59 \text{ (cases)}
$$

Thus, the required probability is $\frac{59}{125}$. Therefore, the answer is $(B)$.