Number of ways to choose $p,q$ from $\{1,3,5,\cdots,399\}$ such that $q > p$? List $M = \{1,3,5...397,399\}$
$M$ contains all positive odd integers from $1$ to $400$, this is not exact wording of the question but basically ask how many ways are there if we randomly choose two numbers $p$ and $q$ from $M$ such that $q > p$?
The answer is $19900$.
My approach was that we know there are $200$ odd integers within $400$, if we randomly picked two numbers, then it must either $q < p$ or $q > p$, so looks like we have $\frac{200\choose2}{2}$, dividing $2$ here means we rule out the case $q < p$.
But I precisely found that $200\choose2$ = $19900$ which is the answer.
I'm a bit confused here then, any suggestion?
 A: You can pick any distinct pair, which you can do in $200 \choose 2$ ways.  You then call the larger one $q$ and the smaller one $p$.  You didn't already call them $p$ and $q$ when you picked them, so you do not divide by $2$.
A: If the first number picked was $1$, there are no winning moves.
If the first number picked was $3$, there is one winning move, namely $1$.
In general, if the first number picked was $2k+1$, there are $k$ winning moves.
$k$ ranges from $0$ to $199$, and $0+1+\dots+199=\binom{200}{2}$.
A: Alternative explanation

if we randomly picked two numbers,...

There are $(200 \times 199)$ ways of randomly picking two numbers.  This is what goes in the numerator of your analysis, re the denominator $= 2.$
With this formula (in the numerator) for picking two random numbers, any pair of distinct random numbers $(q,p)$ will be counted twice, once with $q$ picked first, and once with $p$ picked first.
A: 
if we randomly picked two numbers, then it must either q < p or q > p

Um,.... what if $p =q$?  Isn't that possible?

dividing 2 here means we rule out the case q<p

Why do you think that there are ${200 \choose 2}$ pairs?  Why not $200^2$ pairs.  Or $200 P 2$ pairs.  Or any other number?
========
${200 \choose 2}$ are number of pairs regardless of position.  But you are chooses specific $p$s and $q$s with regard to order.  SO there are $200 P 2  = \frac {200!}{198!} = 200\cdot 199$ and there are not ${200\choose 2}=\frac {200!}{198!2!}$ ways to choose an ordered (but unequal) pair.
Now a nasty hard-nosed professor will ask you to justify that exactly half of the pairs have $p < q$ and exactly half of the pairs have $p > q$.  Couldn't it be possible that there are more pairs where $p < q$ then there are pairs where $q > q$?  Or fewer?
=======
Alternatively.  There are $200^2$ pairs of of $(p,q)$ ordered pairs.
Either $p < q$.  Or $p = q$.  Or $p > q$.
If $p = q$ then $p$ may be $1,3....,399$ so there are $200$ pairs where $p=q$.
So there are $20^2 - 200 = 200(200-1) =200\cdot 199$ pairs $(p,q)$ where $p \ne q$.
And for every pair $(p,q)$ where the first term is less than the second,  we can map it to a pair $(\rho, \tau) = (q, p)$ where the first term is greater than the second.  Likewise for ever pair $(p,q)$ where the first term is greater we can map it to a pair $(\rho,\tau) = (q,p)$ there the first term is smaller.
So pairs where the first term is less, are in one-to-one cooresponde to the pairs where the first term is more.  So the number of pairs $(p,q)$ where $p < q$ are half of all the pairs $(p,q)$ where $p \ne q$.
So there are $\frac {200\cdot 199}2$ such pairs.
======
Third alternative.  If the second term is $2k+1$ and the first term is $< 2k+1$ then the first term can be anything between $1=2\cdot 1-1$ to $2k-1$.  There are $k$ such pairs.
So the total numbers of such pairs is $\sum_{k=1}^{199} k =\frac {199\cdot 200}2$.
