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$\begingroup$ Thanks @knrumsey for showing me another perspective. Your answer does look like a rigorous proof but unfortunately it is too complex for me (not strong in Maths) to follow. I'm not sure what |S| is (absolute value of S?) and where the equality (that follows |S|) comes from. Also, I'm confused about the switching positions of n and k, that is, n choose k (n k) then k choose n (k n). I'll keep going back to your post hopefully one day I will understand it though. $\endgroup$
– Nemo
Commented Nov 18, 2019 at 12:05
1

$\begingroup$ @Nemo |S| denotes the number of elements in set S. The (k,n) switch is a typo. And, you need to look up binomial theorem, i.e. $(x+y)^n$ to understand this better. $\endgroup$
– gunes
Commented Nov 18, 2019 at 12:29
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$\begingroup$ This is not "more rigorous" than the answer from @gunes . It just proves a stronger result (which the OP does not need) and then deduces the size of the power set. $\endgroup$
– Ethan Bolker
Commented Nov 18, 2019 at 16:53
1

$\begingroup$ @Nemo I don't want to discourage you but if this is too complex for you then you may want to consider going back and brushing up on your basics before attacking Casella and Berger. While sigma algebra stuff doesn't play too much of a role past the first introductory chapter (unless they've changed that since I've gone through the book) the arguments laid out in this answer should be something you can follow if you want to get through the chapter on discrete distributions. $\endgroup$
– Dason
Commented Nov 18, 2019 at 21:44
$\begingroup$ Thanks, @Dason. That's a great idea. Could you please advise what book would quickly bring me up to speed of that of Casella and Berger? $\endgroup$
– Nemo
Commented Nov 18, 2019 at 23:22

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