Problem (Combinatorics):
Let set $S=\{1,2,3,4,5,6\}$, and let set $T$ be the set of all subsets of $S$ (including the empty set and $S$ itself). Let $t_{1}, t_{2}, t_{3}$ be elements of $T$, not necessarily distinct. The ordered triple $\left(t_{1}, t_{2}, t_{3}\right)$ is called satisfactory if either

(a) both $t_{1} \subseteq t_{3}$ and $t_{2} \subseteq t_{3}$, or

(b) $t_{3} \subseteq t_{1}$ and $t_{3} \subseteq t_{2}$.

Compute the number of satisfactory ordered triples $\left(t_{1}, t_{2}, t_{3}\right)$.

Solution:
Let $T_{1}=\left\{\left(t_{1}, t_{2}, t_{3}\right) \mid t_{1} \subseteq t_{3}\right.$ and $\left.t_{2} \subseteq t_{3}\right\}$ and let $T_{2}=\left\{\left(t_{1}, t_{2}, t_{3}\right) \mid t_{3} \subseteq t_{1}\right.$ and $\left.t_{3} \subseteq t_{2}\right\}$. Notice that if $\left(t_{1}, t_{2}, t_{3}\right) \in T_{1}$, then $\left(S \backslash t_{1}, S \backslash t_{2}, S \backslash t_{3}\right) \in T_{2}$, so that $\left|T_{1}\right|=\left|T_{2}\right|$. To count $T_{1}$, note that if $t_{1} \subseteq t_{3}$ and $t_{2} \subseteq t_{3}$, then $t_{1} \cup t_{2} \subseteq t_{3}$. Now each set $t_{3}$ has $2^{\left|t_{3}\right|}$ subsets; $t_{1}$ and $t_{2}$ could be any of these, for a total of $\left(2^{\left|t_{3}\right|}\right)^{2}=4^{\left|t_{3}\right|}$ possibilities given a particular subset $t_{3}$. For $n=0,1, \ldots, 6$, if $\left|t_{3}\right|=n$, there are $\left(\begin{array}{l}6 \\ n\end{array}\right)$ choices for the elements of $t_{3}$. So the total number of elements in $T_{1}$ is

$$
\begin{aligned}
\left|T_{1}\right| & =\sum_{k=0}^{6}\left(\begin{array}{l}
6 \\
k
\end{array}\right) 4^{k} \\
& =(4+1)^{6}=15625
\end{aligned}
$$

by the Binomial Theorem. However, $T_{1} \cap T_{2} \neq \emptyset$, because if $t_{1}=t_{2}=t_{3}$, the triple $\left(t_{1}, t_{2}, t_{3}\right)$ satisfies both conditions and is in both sets. Therefore there are 64 triples that are counted in both sets. So $\left|T_{1} \cup T_{2}\right|=2 \cdot 15625-64=\mathbf{3 1 1 8 6}$.

Let $T_{1}$ and $T_{2}$ be defined as above. Then count $\left|T_{1}\right|$ based on the number $n$ of elements in $t_{1} \cup t_{2}$. There are $\left(\begin{array}{l}6 \\ n\end{array}\right)$ ways to choose those $n$ elements. For each element $a$ in $t_{1} \cup t_{2}$, there are three possibilities: $a \in t_{1}$ but not $t_{2}$, or $a \in t_{2}$ but not $t_{1}$, or $a \in t_{1} \cap t_{2}$. Then for each element $b$ in $S \backslash\left(t_{1} \cup t_{2}\right)$, there are two possibilities: either $b \in t_{3}$, or $b \notin t_{3}$. Combine these observations in the table below:

| $\left\|t_{1} \cup t_{2}\right\|$ | Choices for <br> $t_{1} \cup t_{2}$ | Ways of dividing <br> between $t_{1}$ and $t_{2}$ | $\left\|S \backslash\left(t_{1} \cup t_{2}\right)\right\|$ | Choices for $t_{3}$ | Total |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 1 | 1 | 6 | $2^{6}$ | 64 |
| 1 | 6 | 3 | 5 | $2^{5}$ | 576 |
| 2 | 15 | $3^{2}$ | 4 | $2^{4}$ | 2160 |
| 3 | 20 | $3^{3}$ | 3 | $2^{3}$ | 4320 |
| 4 | 15 | $3^{4}$ | 2 | $2^{2}$ | 4860 |
| 5 | 6 | $3^{5}$ | 1 | $2^{1}$ | 2916 |
| 6 | 1 | $3^{6}$ | 0 | $2^{0}$ | 729 |

The total is 15625 , so $\left|T_{1}\right|=\left|T_{2}\right|=15625$. As noted in the first solution, there are 64 triples that are counted in both $T_{1}$ and $T_{2}$, so $\left|T_{1} \cup T_{2}\right|=2 \cdot 15625-64=\mathbf{3 1 1 8 6}$.

Final answer: 31186