<s>[INST] Explain why twice the sum $\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$ is $2^{12}-\binom{12}6$ Can someone explain how is the RHS concluded? I did with sample numbers and it is all correct. but I can't figure out how C(12,6) comes to play.
$$
\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} = (2^{12} - \binom{12}{6}) / 2
$$
 [/INST] Hint: We have
$$
\sum_{i=0}^n \binom{n}{i} = 2^n
$$
and
$$
\binom{n}{i} = \binom{n}{n-i}
$$
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