Binomial theorem incomplete expansion I got this question and I am little bit confused, whether the question is correct or not.
$n\choose0$-$ n\choose1 $+$ n\choose2$-$ n\choose3$+.........+$(-1)^r$$  n\choose r$=$28$

Now we are required to find the values of n. Its obvious that its an incomplete expression of $(1-x)^n$ but if we don't know the $r$ how can we go on to find $n$. The answer was $n=9$ but actually it would be worthless if we don't know the values where we have to stop i.e the value of $r$.

 A: That's because
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,r} {\left( { - 1} \right)^{\,k} \left( \matrix{
  n \cr 
  k \cr}  \right)}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  r - k \cr 
  r - k \cr}  \right)\left( \matrix{
  n \cr 
  k \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,r} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( \matrix{
   - 1 \cr 
  r - k \cr}  \right)\left( \matrix{
  n \cr 
  k \cr}  \right)}  = \left( { - 1} \right)^{\,r} \left( \matrix{
  n - 1 \cr 
  r \cr}  \right) \cr} 
$$
after which
$$
28 = 7 \cdot 4 = {{8 \cdot 7} \over {1 \cdot 2}}\quad  \to \quad \left( {n = 9,\;r = 2} \right)\; \vee \;\left( {n = 9,\;r = 6} \right)
$$
A: The expression $\sum_{k=0}^{r}{n \choose k}{(-1)}^k$ can be simplified as ${(-1)}^r{n-r \choose r}$. Skipping $r=0,1$, you can see that $r$ is even and for $r \geq 4$ there is no solution, alternatively see where 28 appears in Pascal's triangle.
