# A new method!

Through this note,I seek to present to the Brilliant community a new way of proving that $$\displaystyle \sum_{r=0}^{r=n}\dbinom{n}{r}=2^n$$. $$\text{First Method}$$: We already know this way,$(1+x)^n=\sum_{r=0}^{n}\dbinom{n}{r}x^r\\ \Longrightarrow (1+1)^n=2^n=\sum_{r=0}^{n}\dbinom{n}{r}$,which is what we wanted to prove.$$\text{Second Method}:$$Consider $$n$$ identical coins.Each coin has two faces,Heads and Tails,now we count the number of ways in which their faces can be arranged,in two different ways: $$\text{First Way:Rule of Product,number of ways}=2^n\\ \text{Second Way:}$$

Let us say that in the $$r^{th}$$ arrangement there are $$r$$ heads,so number of arrangements=$$\dbinom{n}{r}$$.Here $$0 \leq r \leq n$$,hence total number of ways$=\sum_{r=0}^{r=n}\dbinom{n}{r}$,hence we get $\sum_{r=0}^{n}\dbinom{n}{r}=2^n$.And done!

2 years, 4 months ago

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