A box contains 10 pairs of shoes (20 shoes in total). If two

You are right it's not the same thing:

First of all:
In numerator you have the number of selection of ONE shoe out of 20, but in denominator TWO shoes out of 20. In my solution both numerator and denominator counting the # of selections of TWO shoes, though from different quantities.

Second, consider this: what is your winning (favorable) probability? You want to choose ONE pair. Out of how many pairs? Out of 10. Now, what are you actually doing? You are taking TWO shoes out of 20, "hoping" that this two will be the pair. So anyway wining scenarios are 10C2 and total number of scenarios are 20C2.

You can do this also in the following way (maybe this one would be easier to understand):
Numerator: 20C1 - # of selection of 1 shoe out of 20, multiplied by 1C1, as the second one can be only ONE, as there is only one pair of chosen one. Which means that # of selection would be 1C1. Put this in numerator.

Denominator would be: 20C1 # of selection of 1 from 20, multiplied be 19C1 # of selection from the 19 left.

So, P=20C1*1C1/20C1*19C1=1/19, here we have in the numerator the same thing you wrote, BUT if we are doing this way then in denominator you should also count the # of selection of the first one out of 20 and the second one out of 19.

The first way of solving actually is the same. Take one shoe from 20, any shoe from 20, I mean just randomly take one. Then you are looking at your 19 left shoes and want to choose the pair of the one you've already taken, as in 19 you have only one which is the pair of the first one you have the probability 1/19 (1 chance out of 19) to choose the right one.