For how many integers $n$ where $2 \le n \le 100$ is $\binom{n}{2}$ odd? For how many integers $n$ where $2 \le n \le 100$ is $\binom{n}{2}$ odd?
I'm finding lots of odd results, but no pattern yet. Any hints are greatlyappreciated.
 A: Let's write explicitly: $\binom{n}{2} = \frac{n(n-1)}{2}$ and that is odd exactly if $4$ does not divide neither $n$, $n-1$ (obviously it doesn't divide both and neither does $2$). Which means $n\equiv 3$ or $2$  (mod  $4$). Now count.
A: Hint: draw Pascal's triangle mod 2:
1
1 1
1 0 1
1 1 1 1
1 0 0 0 1
1 1 0 0 1 1
1 0 1 0 1 0 1
1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0 1 1
1 0 1 0 0 0 0 0 1 0 1
1 1 1 1 0 0 0 0 1 1 1 1

etc.
A: $$\binom{a}{b} = \frac{a!}{b! * (a - b)!}$$
Using that we should be able to determine which are odd. The results to your formula will always be 
$$f(n) = \frac{n!}{2 *(n-2)!}=\frac{1}{2} * \frac{n!}{(n - 2)!}$$ 
Because it has the $\frac{1}{2}$ factor, it indeed will have many odds. Also note that $\frac{n!}{(n-2)!}$ just means $n * (n - 1) = n^2 - n$ so the formula actually is 
$$f(n) = \frac{1}{2}n^2 - \frac{1}{2}n$$
Now you can make a nice graph of this and look up for $D_f = [2, 100]$ which results are even.
EDT: Of course you could also create a new formula like 
$$g(n) = (\frac{1}{2}n^2 - \frac{1}{2}n) \mod{2}$$
from which the odd numbers can be recognized more easily (just count how often $g(n) = 0$)
