Number of way choose ordered distinct quadruples $(A,B,C,D)$ from $\{1,2,...,n\}(n\geq 2)$ 
Number of way choose ordered distinct quadruples $(A,B,C,D)$ from
$\{1,2,...,n\}(n\geq 2)$ such that  $A\subseteq B\cup C\cup D$.

I think in following manner:
$\forall x\in \{1,2,...,n\} $
if $A$ included it then it must be in  $B \vee C \vee D$ or if $x$ not in $A$ it can be at any $B \vee C \vee D$ such a way give us $7^n$ .Any one can verify my solution?

Let's count the $(A,B,C,D)$ for a given choice of $S=B\cup C\cup D$ (where $k=|S|$).
Draw a Venn diagram for $B,C,D$. We can distribute the elements of $S$ among the $7$ regions, and that corresponds to a choice of $(B,C,D)$. Labelling the regions $1$-$7$, this is the same as a function from $S$ to the set $\{1,\cdots,7\}$, of which there are $7^k$. Also there are $2^k$ choices for $A\subseteq S$.
There are $\binom{n}{k}$ choices for $S\subseteq\{1,\cdots,n\}$, so summing over $k$ we get
$$ \sum_{k=0}^n \binom{n}{k}\cdot7^k\cdot2^k. $$
Do you see why this is $15^n$?
