# Relating variables: $A_{i}$ with $|A_{i}|=p$ and $B_{j}$ with $|B_{j}|=q$ such that $\cup_{i=1}^{n}A_{i}=\cup_{j=1}^{m}B_{j}=S$ with constraint on $S$

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.

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.

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

$$A_i$$'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 $$|S| = \dfrac{mp}{\alpha} = \dfrac{nq}{\beta}$$

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