Binomial Expansion related problem If   ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^r}{}^n{C_r} = 28$. Then find the value of $n$.
My approach is as follow ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^r}{}^n{C_r} = 28 = {}^8{C_2} = {}^8{C_6}$
${\left( {1 - x} \right)^n} = {}^n{C_0} - {}^n{C_1}x + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} + .. + {\left( { - 1} \right)^n}{}^n{C_n}.{x^n}$
Putting x=1
${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^n}{}^n{C_n} = 0$
$ \Rightarrow 28 + {\left( { - 1} \right)^{r + 1}}{}^n{C_{r + 1}} + .. + {\left( { - 1} \right)^n}{}^n{C_n} = 0$
How do I approach from here

Pascal's identity is sufficient :
$$ {N-1 \choose R-1} + {N-1 \choose R} = {N \choose R}$$
Since $${n \choose 0} = 1 = {n-1 \choose 0}$$
given expression is
$${n-1 \choose 0} - {n \choose 1} + {n \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
Operating on first two terms every step,
$$= \color{red}{- {n-1 \choose 1}} + {n \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
$$= \color{blue}{ n-1 \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
$$ \cdots $$
$$= \color{brown}{(-1)^r{n-1 \choose r}}$$
Thus as you have identified, $n-1 = 8$ and $r=2$ or $6$.
