<s>[INST] Prove that the sequence of combinations contains an odd number of odd numbers 
Let $n$ be an odd integer more than one. Prove that the sequence $$\binom{n}{1}, \binom{n}{2}, \ldots,\binom{n}{\frac{n-1}{2}}$$
  contains an odd number of odd numbers.

I tried writing out the combination form as $$\frac{(2k+1)!}{(m!)((2k+1)-m)!}.$$ How do I use this to show that the sequence contains an odd number of odd numbers?
 [/INST] $$
\overbrace{\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2}} + \overbrace{\binom n {\frac{n+1} 2} + \cdots + \binom n {n-1} + \binom n n} = 2^n.
$$
The two sums under the $\overbrace{\text{overbraces}}$ are equal; hence
$$
\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2} = 2^{n-1}.
$$
Therefore
$$
\binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2} = 2^{n-1}-1 = \text{an odd number}.
$$
If a sum of finitely many terms is an odd number (as this one is) then the number of odd terms must be odd, since if it were even, then the sum would be even.
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