<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] Suppose we toss a fair coin for $12$ times. The number of combinations is $2^{12}$.
Now notice the number of outcomes $N(H>T)$ such that the total number of Head is more than Tail is same as that of Tail more than Head $N(H<T)$, by symmetry.
We know $N(H>T)$ is the total number of choosing less than $6$ Tails, which is 
$$N=\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$$
We know that the no. of outcomes when the number of Head and Tail equals is ${12}\choose{6}$. 
Since the total number of possible outcome is 
$$2N+{{12}\choose{6}}=2^{12}$$
The claim follows.
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