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?
Best Answer
So we have $$\sum _{k=0}^n{n\choose k}\cdot 7^k\cdot 8^{n-k} = 15^n$$ 4-couples $(A,B,C,D)$.