Simplify: $$\frac{1}{P(1-\frac{P}{N})(1-\frac{m}{P})}$$ To show that it is equal to:
$$\frac{N}{N-m} \left[\frac{1}{N-P}+\frac{1}{P-m}\right].$$
I honestly have no idea how to even start the question. I am wondering if the question is asking to go backward, i.e. to go from the second part to the first, but it doesn't seem that way. If someone could show me how to do this or even get me started would be very helpful!
Hint:
$$\frac{1}{P(1-\frac{P}{N})(1-\frac{m}{P})}=\frac{1}{(1-\frac{P}{N})(P-m)}=\frac{N}{(N-P)(P-m)}.$$
Then use partial fractions.
$$\frac{N}{(N-P)(P-m)}=\frac{\alpha}{N-P}+\frac{\beta}{P-m}$$
Expand the expression on the right and compare it with the expression on the left-hand side to determine $\alpha$ and $\beta$.
Euler's trick: Instead of expanding the right-hand side you can also use Euler's trick. For example to obtain $\alpha$ multiply by the associated denominator $N-P$ and then set $N-P=0\implies P=N$ to obtain:
$$\frac{N}{P-m}\biggl|_{P=N}=\alpha+\frac{N-P}{P-m} \biggl|_{P=N}\beta \implies \alpha = \frac{N}{N-m}$$