Question

simplify C(n,1)+C(n,2)+C(n,3)+... +C(n,n-1)+C(n,n)
where C(n,n)=n!/(r!(n-r)!)

Answer 1

It's like the binomial theorem

Answer 2

C(n,r)=C(n,n-r)

Answer 3

\[ \Large
(a+b)^n = \sum_{k=0}^n\binom{n}{k}a^{n-k}b^k
\]

Answer 4

Let \(a=1, b=1\) and you get: \[
\Large
2^n = \sum_{k=0}^n\binom{n}{k}
\]

Answer 5

\[
\Large
2^n = \sum_{k=0}^n\binom{n}{k} = \sum_{k=1}^n\binom{n}{k}+\binom{n}{1}
\]This gives me\[ \Large
\sum_{k=1}^n\binom{n}{k} =2^n- \binom{n}{1}
\]

Answer 6

@bernlim Make sense?

Answer 7

yeah, thanks!