Finding the limit of ${n-1 \choose k}p^k(1-p)^{n-1-k}$ as $n$ goes to infinity

Question

On this Wikipedia page on random graphs, they compute this limit to be

$$\lim_{n\to\infty}\binom{n-1}kp^k(1-p)^{n-1-k}=\frac{(np)^ke^{-np}}{k!}$$

with $np$ constant. Any hints on how to get that? Where does the $e$ come from?

NOTE: Hints rather than complete solutions would be appreciated.

Answer 1

Since $k$ is held constant and $n\to\infty$, the binomial coefficient is $1+o(1)$ times $n^k/k!$, and the $e$ arises in the usual way from $(1-p)^{1/p}$.

Thinking probabilistically, this can be rephrased in terms of Poisson approximation.

Answer 2

Any hint will be an almost complete solution: $$ =\frac{(1-\frac1n)·…·(1-\frac kn)}{k!}·(np)^k·\left(1-\frac{(np)}{n}\right)^{n-(k+1)} $$ all factors converge under the assumption that $k$ and $np$ are constant.

Answer 3

Hint: To keep it a hint fill in the dots: $$(1-p)^n=\sum_{k=0}^{n}\dbinom{n}{k}(-p)^k=\ldots\approx\sum_{k=0}^n\frac{(-np)^k}{k!}$$ and $$\dbinom{n}{k}=\ldots\approx \frac{n^k}{k!}$$

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You can find more about this under the search: "limit of binomial distribution to Poisson distribution" (Of course $n\to +\infty$ implies that $n-1 \to +\infty$, so I simplified the $n-1$).