What is a method to compute the asymptotics of a sum resulting from the inclusion-exclusion principle? Each term of the sum can be approximated perhaps by Stirling's formula or the Gaussian distribution. However the alternating sign should effect some cancellation. As an example, the [answer to this combinatorial problem](https://math.stackexchange.com/a/3172541/64809) generates a probability
$$p=\frac c{n\choose j},$$
where
$$c=\sum\_{k=w}^j(-1)^{k-w}(j-k+1){n-k+1\choose j-k+1}.$$
What is the asymptote of $p$ for $\big|\frac jn-a\big|=o(n)$ and $\big|\frac wn-b\big|=o(n)$ for some positive numbers $a$ and $b$ as $n\rightarrow\infty$?



Asymptotics for the sums from the inclusion-exclusion principle
Another way to approach this from the generating function perspective. Notice that
$$\sum\_{k=l}^u (-1)^k [x^k]\, f(x) = [x^u]\,\frac{f(-x)}{1-x} - [x^{l-1}]\,\frac{f(-x)}{1-x},$$
and so the question reduces to studying the asymptotic of the coefficients of $\frac{f(-x)}{1-x}$.


In your example, we have 
$$(j-k+1)\binom{n-k+1}{j-k+1} = (n-j+1)\,[x^{j-k}]\,(1-x)^{-(n+2-j)}$$
and thus
\begin{split}
c &=\sum\_{k=w}^{j} (-1)^{k-w} (j-k+1)\binom{n-k+1}{j-k+1} \\
&= (n-j+1)(-1)^{j-w} \sum\_{t=0}^{j-w} (-1)^t\,[x^t]\,(1-x)^{-(n+2-j)} \\
&= (n-j+1)(-1)^{j-w}\,[x^{j-w}]\,\frac{1}{(1+x)^{n+2-j}(1-x)} \\
&= (n-j+1)\,[x^{j-w}]\,\frac{1}{(1-x)^{n+2-j}(1+x)}.
\end{split}


Now, noticing the [partial fraction decomposition](https://en.wikipedia.org/wiki/Partial_fraction_decomposition)
$$\frac{1}{(1-x)^m(1+x)} = \frac{2^{-m}}{1+x} + \sum\_{i=0}^{m-1} \frac{2^{-(i+1)}}{(1-x)^{m-i}},$$
we obtain the formula:
$$c=(n-j+1)\left(\frac{(-1)^{j-w}}{2^{n+2-j}} + \sum\_{i=0}^{n+1-j} \frac{1}{2^{i+1}}\binom{n+1-i-w}{j-w}\right).$$
Here all terms in the sum are positive and thus are more amenable to asymptotic analysis. For example, we can deduce
$$\frac{n-j+1}{2}\binom{n+1-w}{j-w} \lesssim c \lesssim (n-j+1)\binom{n+1-w}{j-w}.$$


