Consider two disjoint sets $A$ and $B$ with $|A|=|B|=8$. Find two ways to count the number of ordered pairs $(X,Y)$ with $X \subseteq A$ and $Y \subseteq (B \cup X)$ with $|Y|=8$.

Is my guess correct?

  • We have $|B \cup X| = |B| + |X| - |B \cap X| = 8+|X|$ (since $A$ and $B$ are disjoint, $X \subseteq A$). We know that $1 \le^* |X| \le |A| = 8$ and $|Y| = 8 \le 8 + |X|$.

(*We'll leave out $|X| = 0$, since it does not contribute to the number of ordered pairs $(X,Y)$).

  • This results in $ 16 \ge |B \cup X| \ge 9 $. So for all possible elements for $Y$ we do the following: $\sum_{i=9}^{16} \binom{i}{8}$. There are from $1$ through $8$ possible elements for $X$ (chosen from $A$): $\sum_{k=1}^8 \binom{8}{k}$.

  • The number of ordered pairs is $\sum_{k=1}^8 \binom{8}{k} \cdot \sum_{i=9}^{16} \binom{i}{8}$.

I'm struggling to find another way.

  • $\begingroup$ "couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair". $\endgroup$ – Jean Marie Jan 28 at 19:43
  • $\begingroup$ Thanks! I edited my post. Do you think my answer is correct, @JeanMarie? $\endgroup$ – Zachary Jan 28 at 19:49
  • 3
    $\begingroup$ If I compute well, the number of such ordered pairs should be $\sum_{k=\color{red}{0}}^8 \binom{8}{k} \cdot \binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation. $\endgroup$ – Jean Marie Jan 28 at 20:06

Formula: #(A×B) = #A × #B
For each a in A there are #B ways to make a into an ordered pair.
Thus add #B, #A times.

The other way.
For each b in B, there are #A ways to make b ...


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