13. Determine the number of 4 -element subsets $\{a, b, c, d\}$ of $\{1,2,3,4, \cdots, 20\}$ such that $a+b+c+d$ is divisible by 3 .
13. Answer: 11901 .
For $i=0,1,2$, let
$$
A_{i}=\{j: 1 \leq j \leq 20, j \equiv i \quad(\text { od } 3) \quad\} .
$$

Note that $\left|A_{0}\right|=6,\left|A_{1}\right|=7$ and $\left|A_{2}\right|=7$.
It can be shown that for $a, b, c, d \in\{1,2,3,4, \cdots, 20\}, a+b+c+d$ is divisible by 3 iff
(i) $\left|\{a, b, c, d\} \cap A_{i}\right|=2$ for all $i=1,2$; or
(ii) $\left|\{a, b, c, d\} \cap A_{0}\right|=1$ and $\left|\{a, b, c, d\} \cap A_{i}\right|=3$ for some $i: 1 \leq i \leq 2$; or
(iii) $\left|\{a, b, c, d\} \cap A_{0}\right|=2$ and $\left|\{a, b, c, d\} \cap A_{i}\right|=1$ for all $i=1,2$; or
(iv) $\left|\{a, b, c, d\} \cap A_{0}\right|=4$.

Thus the number of 4 -element subsets $\{a, b, c, d\}$ of $\{1,2,3,4, \cdots, 20\}$ such that $a+b+$ $c+d$ is divisible by 3 is
$$
\begin{aligned}
\binom{\left|A_{1}\right|}{2}\binom{\left|A_{2}\right|}{2}+ & \binom{\left|A_{0}\right|}{1}\binom{\left|A_{1}\right|}{3}+\binom{\left|A_{0}\right|}{1}\binom{\left|A_{2}\right|}{3}+\binom{\left|A_{0}\right|}{2} \times\left|A_{1}\right| \times\left|A_{2}\right|+\binom{\left|A_{0}\right|}{4} \\
& =\binom{7}{2}^{2}+6\binom{7}{3} \times 2+\binom{6}{2}^{2} \times 7^{2}+\binom{6}{4}=11901 .
\end{aligned}
$$