Summation of Binomial Theorem

Question

The binomial theorem formula: $$\sum\limits_{k=0}^{n} {n \choose k} = \sum\limits_{k=0}^{n}\frac{n!}{k!(n-k)!} = \sum\limits_{k=0}^{n}\frac{n(n-1)(n-2) \cdots (n-k+1)}{k(k-1) \cdots 2\cdot1}.$$

I am having a difficult time getting from $\displaystyle\frac{n!}{k!(n-k)!}$ to $\displaystyle\frac{n(n-1)(n-2) \cdots (n-k+1)}{k(k-1) \cdots 2 \cdot 1}$.

I am trying to use this equality but I cannot show it to myself. I have tried to input values of $k$ to see what it looks like, but I am still having trouble.

Any hint would be greatly appreciated.

Answer 1

$$\frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2) \cdots (n-k+1) (n-k)!}{k!(n-k)!} = \frac{n(n-1)(n-2) \cdots (n-k+1)}{k!}$$

For instance, if $n=7$ and $k=4$, then $$\displaystyle{7 \choose 4} = \frac{7!}{4!\cdot(7-4)!} = \frac{7!}{4!\cdot3!} = \frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{4!\cdot3\cdot2\cdot1} = \frac{7\cdot6\cdot5\cdot4}{4!}.$$

Answer 2

The sum of the binomial coefficients is in fact $2^n$. What you quoted was the general formula for the binomial coefficient. You need to multiply top and bottom by $(n-k)!$, from which the result comes immediately.