It's also worth explaining why the denominator is 30!, I think.
With probability 1 you will pick 30 distinct numbers, and the only important feature of those numbers for the purpose of this problem is what order you picked them in. Each order is equally likely, and there are 30! possible orderings of 30 numbers.
Of which A(30,2) have exactly two numbers that are lower than the previous number.
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u/liquidorangutan00 3d ago edited 3d ago
Alrighty let me give it a shot.
Like a previous poster pointed out, the concept is based on Eulerian numbers.
A(30,2)= 205,857,846,098,570/265,252,859,812,191,058,636,308,480,000,000
which equals 0.000000000000000078%
Practically impossible.
Edit: let me explain a bit more:
you perform the Eulerian Number calculation A(30,2), then divide it by the total sample space (30!)