I've been trying to solve this combinatorial problem but couldn't solve it. The problem is as follows: If C(a,b) is the binomial coefficient then prove that $$ C(n,0) + C(n,2) + C(n,4) +...+C(n,n) = C(n,1)+C(n,3)+C(n,5)+...+C(n,n-1)=\frac{2^n}{2}$$ I would really appreciate a step by step solution or at least a hint on how to solve it because I've been trying for days.
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yes, I don't know how to write the binomial coefficient. I will edit the name – Mr. Nicolas Mar 16 '23 at 09:34
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2Hint. $(1+1)^n=2^n$, $(1-1)^n=0$. – Robert Z Mar 16 '23 at 09:36
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Think about counting subsets. – Ben Mar 16 '23 at 09:36
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Thanks for the hint. I tried by writing 2^(n-1) as a sum. Then I got a sum with binomial coefficient but couldn't write it as the other sums – Mr. Nicolas Mar 16 '23 at 09:39
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Probably been asked and answered here many times. I recommend a search. – Gerry Myerson Mar 16 '23 at 09:49
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Call $$x= C(n,0) + \dots + C(n,n)$$ and $$y= C(n,1) + \dots + C(n,n-1)$$ Then $$x+y = \sum_{k=0}^n C(n,k)=2^n$$ and $$x-y= \sum_{k=0}^n C(n,k) (-1)^k =0$$
Hence $x,y$ are the solutions of the system $$\cases{ x+y=2^n \\ x-y=0}$$ which can be easily solved.
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1$(a+b)^n=\sum_{k=0}^n C(n,k) a^k b^{n-k}$. Set $a=-1$, $b=1$. The solution and the question obviously presume that $n$ is even. – van der Wolf Mar 16 '23 at 09:53
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