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. $$ \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} = (2^{12} - \binom{12}{6}) / 2 $$
Asked
Active
Viewed 60 times
1
-
2Do you know the value of $$C(12,0) + C(12,1) + ... + C(12,12)$$ ?? – quasi Mar 22 '17 at 08:57
-
Yes, it is. C(12,0)+C(12,1)+...+C(12,12) = 2^12 – Sanone Mar 22 '17 at 09:04
-
Right. Now using the hint in Yikai's answer, if $$ x = C(12,0) + ... + C(12,5)$$ $$ y = C(12,12) + ... + C(12,7)$$ how do $x,y$ compare? – quasi Mar 22 '17 at 09:12
-
I see that x = y (I am not really good in maths :( ) – Sanone Mar 22 '17 at 09:13
-
So, if you subtract $C(12,6)$ from $2^{12}$, what's left? – quasi Mar 22 '17 at 09:14
-
When I subtract C(12,6), I have 2x (or 2y)s. – Sanone Mar 22 '17 at 09:16
-
So, now you have it. Done! – quasi Mar 22 '17 at 09:16
2 Answers
3
Hint: We have $$ \sum_{i=0}^n \binom{n}{i} = 2^n $$ and $$ \binom{n}{i} = \binom{n}{n-i} $$
PSPACEhard
- 10,283
0
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.
Mythomorphic
- 6,008