Alternating sum of binomial of coefficients up to $\binom{n}{r}$ Question : Evaluate - $$ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r}$$
I tried to solve it but I am able to solve only when $r=n$.In that case I can put $x=-1$ in the expansion of $(1+x)^n$ and the result is $0$.
I know that for asking any question on this forum, I need to show my approach and thoughts but literally I have no idea how to solve up to $r+1$ terms.
Although I tried to add and subtract the terms to reach upto $\binom{n}{n}$, and evaluate the remaining sum, but nothing helpful.
Any help will be appreciable!

I also apply:
$$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$
But in another way, call:
$$f(n) = \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r} = \sum\limits_{k=0}^{r} \binom{n}{k}-1^k$$
$$f(n+1)-f(n)=\sum\limits_{k=0}^{r} -1^k (\binom{n+1}{k}-\binom{n}{k})=\sum\limits_{k=0}^{r} -1^k \binom{n}{k-1}=\sum\limits_{k=0}^{r-1} -1^{k+1} \binom{n}{k}$$  
$$f(n+1)=f(n+1)-f(n)+f(n)=\sum\limits_{k=0}^{r-1} -1^{k+1} \binom{n}{k} + \sum\limits_{k=0}^{r} \binom{n}{k}-1^k = -1^r\binom{n}{r} +  \sum\limits_{k=0}^{r-1} (-1^k+(-1^{k+1})) \binom{n}{k} = -1^r\binom{n}{r}$$
Because
$$f(n+1) = -1^r\binom{n}{r}$$
So
$$f(n) = -1^r\binom{n-1}{r}$$
