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!

#Combinatorics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$