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!

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