Equivalent expression for $\sum_{m=0}^{n}{n \choose n-m}$

Question

I want an equivalent expression for this expression $\sum_{m=0}^{n}{n \choose n-m}$.

I expanded getting $\sum_{m=0}^{n}{n \choose n-m}=1+n+\frac{n(n-1)}{2!}+ \frac{n(n-1)(n-2)}{3!}+\cdots + n!$

It seems I'm getting a Taylor expansion, which is similar to the expansion of $e^x$

Answer 1

$$\sum_{m = 0}^{n} \binom{n}{n - m} = \sum_{m = 0}^{n} \binom{n}{m} = 2^{n}.$$ You might like to look into the Binomial Theorem and/or Pascal's Triangle.

Answer 2

Hint: Recall that $\binom{n}{n-m} = \frac{n!}{(n-m)!(n-(n-m))!} = \frac{n!}{(n-m)!m!} = \binom{n}{m}$.

So $\sum_{m=0}^n\binom{n}{n-m} = \sum_{m=0}^n\binom{n}{m}$. You can actually simplify this furthur if you'd like.