We usually require $n$ to be a positive integer when defining the combination number $${n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots 1}$$

However, it is possible to relax this restriction and let $n$ to be any real (and even complex) number $\alpha$. In such a case, the definition still stays the same: $${\alpha\choose k}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-k+1)}{k(k-1)(k-2)\cdots 1}$$

With this definition, the binomial expansion can be extended to non-integer power (for concise reason, we use $(1+x)$ instead of $(a+b)$, but the latter also works) $$(1+x)^{\alpha}=\sum_{k=0}^{\infty}{\alpha\choose k}x^{k}$$

It can be verified that the binomial expansion formula is a special case of this one. Note that when $n$ and $k$ both are positive integers, and $k > n$, we have $${n\choose k} =\frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots 1}=0 $$ because one term in the numerator is zero. It follows that $$\sum_{k=0}^{\infty}{n\choose k}x^{k}=\sum_{k=0}^{n}{n\choose k}x^{k}$$

This generalized definition of combination number is not required at high school level competitions. However, some advanced level students may find it useful when solving some challenging problems or learning college level math.