1
$\begingroup$

I am supposed to relate variables in the following problem. It would be great if someone could explain what the problem statement means and also how to tackle such a problem.

enter image description here

I do understand the notation, also, in the question image $O(X)$ denotes the cardinality of set $X$. I have used $|X|$ in the title. At present, I have no idea on how to proceed. Any hints are appreciated. Thanks.

$\endgroup$
  • $\begingroup$ Hint: Double counting. What is $\sum |A_i|$ expressed in 2 different ways? $\endgroup$ – Calvin Lin Oct 26 '20 at 14:26
1
$\begingroup$

$A_i$'s are collection of $m$ sets such that there are $p$ elements in each of them. $B_j$'s are collection of $n$ sets such that there are $q$ elements in each of them.

Any element in a given $A_i$ also occurs in exactly $\alpha -1$ other $A_i$'s $(\alpha \le m)$. Similarly any element in a given $B_j$ also occurs in exactly $\beta -1$ other $A_i$'s $(\beta \le n)$.

Also union of all $A_i$'s is identical with union of all $B_j$'s. This means same total objects make up both the collections, $\{A_i\}$ and $\{B_j\}$. These distinct objects are elements of (universal) set $S$.

Now there are $mp$ elements in total in $\{A_i\}$, each repeated exactly $\alpha$ times. Hence number of distinct elements in $\{A_i\}$ is only $$\dfrac{mp}{\alpha} $$

Similarly, number of distinct elements in $\{B_i\}$ is only $$\dfrac{nq}{\beta} $$

And given is $$|S| = \dfrac{mp}{\alpha} = \dfrac{nq}{\beta}$$

$\endgroup$
  • 1
    $\begingroup$ Thanks for answering. This explains the situation well. Cheers. $\endgroup$ – Paras Khosla Nov 1 '20 at 8:25

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.