Explain why twice the sum $\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$ is $2^{12}-\binom{12}6$

Question

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. $$ \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} = (2^{12} - \binom{12}{6}) / 2 $$

Answer 1

Hint: We have $$ \sum_{i=0}^n \binom{n}{i} = 2^n $$ and $$ \binom{n}{i} = \binom{n}{n-i} $$

Answer 2

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.