I know that
If $np$ is of order of one, then $$P(k_1 ,k_2 , \cdots ,k_m) = \frac{n!}{k_1! k_2! \cdots k_m!}p_1^{k_1}p_2 ^{k_2}\cdots p_{m}^{k_r} \overset{n\to\infty}{\longrightarrow} e^{-np_1}\frac{(np_1)^{k_1}}{k_1!} \cdots e^{-np_m}\frac{(np_m)^{k_m}}{k_m!},$$ where $n=k_1+\cdots+k_m$ and $p_1 +\cdots +p_m =1$.
However, I met the problem that
Under the condition of $k_1 p_1 \ll 1$ and $k_2p_2 \ll 1$, prove that relationship of $$\frac{n!}{k_1!k_2!k_3!} \approx \frac{n^{k_1+k_2}}{k_1! k_2!} \quad\text{and}\quad p_3^{k_3} \approx e^{-n(p_1+p_2)}.$$ and prove the generalized Bernoulli-Trials probability to Poisson probability approximation relation for the three events of occurrence.
The condition is not that $np$is of order of one. How can I prove with the condition of $k_1 p_1 \ll 1$ and $k_2p_2 \ll 1$?
Actually when $n$ is infinitely large, I can prove that $$\frac{n!}{k_3!} \approx n^{k_1}\times n^{k_2}\quad\text{and}\quad p_3^{k_3} \approx e^{-n(p_1+p_2)}.$$