$\begin{aligned} 2020^{2019} & = (2019+1)^{2019} \\ & = 2019^{2019} + 2019 \cdot 2019^{2019} + \frac {2019\cdot 2018}2 \cdot 2019^{2017} + \cdots + 1 \\ & < 3 \cdot 2019^{2019} < 2019^{2020} \end{aligned}$

Therefore $\boxed {2020^{2019} < 2019^{2020}}$ .

Logic Method :

If $x > y$ , then $y^{x} > x^{y}$ as long as $x, y > e(2.718.....)$ .

So $2019^{2020}$ is greater and statement $2$ is right.

$\hspace{1px}$

Python Code :

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a = 2019
b = 2020

if b**a > a**b :
    print("Statement 1 is the answer")
elif b**a < a**b :
    print("Statement 2 is the answer")
elif b**a == a**b :
    print("Statement 3 is the answer")
else :
    print("We can't say")

#The output will be - "Statement 2 is the answer"

For $x>y>e$ , (" $e$ " is the Euler's number : $2<e<3$ ) we have $x^{\frac 1x}<y^{\frac 1y}$ .

So $2020^{\frac {1}{2020}}<2019^{\frac{1}{2019}}$

$\implies \boxed {2020^{2019}<2019^{2020}}$ .