Finding the Number of Sets in Two Groups: A Simplified Problem

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The discussion focuses on a mathematical problem involving two groups of sets, A and B, where the number of sets in group B (denoted as 'n') is expressed in terms of the parameters m, p, p1, q, and q1. The problem is simplified by analyzing the sample space S and counting the elements in the set Ω using two different methods. The conclusion drawn is that |S| p1 = pm, establishing a relationship between the parameters that can be further explored to derive the value of 'n'.

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Came to know about the following problem from a friend which can be simplified to the following:
A1, A2, ...Am and B1, B2,...Bn are two groups of sets each group spanning the sample space.
Now there are p elements in each of Ai and each element is in exactly p1 of the sets of the A group.
Again there are q elements in each of Bi and each element is in exactly q1 of the sets of the B group.
We have to write, 'n' in terms of m,p,p1,q,q1. Thanks for any ideas.
 
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Denote the sample space by ##S##. Let us take the set ##\Omega = \{(x,k)~\vert~x\in A_k\}##. Let's count the elements in ##\Omega## in two ways. In the first way, we first choose an element ##x\in S##, this can be done in ##|S|## ways. Then I choose ##k## such that ##x\in A_k##. This can be done in ##p_1## ways. Furthermore every choice I make yields different elements of ##\Omega##. So ##|\Omega| = |S| p_1##.
I can also choose ##k## first, this can be done in ##m## ways. Then I can choose ##x## such that ##x\in A_k##, this can be done in ##p## ways. Every choice I made yields different elements of ##\Omega##. Thus ##|\Omega| = pm##. So we get ##|S| p_1 = pm##. I leave the rest of the solution to you.
 
Thanks a lot.
 

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