1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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
0
$\begingroup$

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 ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.