TITLE: Probability of selecting distinct numbers in ascending order
QUESTION [2 upvotes]: A relatively simple problem is stumping me on this :(
Given a set of 4 distinct numbers that can be chosen from [0-9], what is the probability that after selecting, the first two numbers would be in ascending order?
Here's my thinking so far:

If we select 0 as our first, we have a 9/10 chance of choosing an appropriate second number
If we select 1 as our first, we have an 8/10 chance of choosing an appropriate second number
...
If we select 8 as our first, we only have a 1/10 chance of choosing an appropriate second number
However if we select 9 as our first, we can't select anything greater

So the probability of selecting the first two numbers such that they are ascending should be:
P(1st selection) * P(2nd selection > 1st selection) * P(3rd selection that we don't care about) * P(4th selection that we don't care about) = 1/10 * 1/9 * 1/8 * 1/7 => 1/5040
I don't think I'm going about this problem the right way and my answer is completely off - I don't know how to handle what the probability of selecting the second number would be, since we could choose a great number like 0 for the first, or a bad one like 8 or 9 that would give us less choices.  Could someone explain this in a better way?

REPLY [0 votes]: Let $A$ denote the event that the first two numbers are ascending and let $D$ denote the event that the first two numbers are descending. These events are disjoint and complements of eachother. 
That means that $$P\left(A\right)+P\left(D\right)=1$$
Next to
that there is no reason to assume that $P\left(A\right)$ differs
from $P\left(D\right)$ so: $$P\left(A\right)=P\left(D\right)$$
(if you would change $0$ into $9$, $1$ into $8$ et cetera, then in essence nothing would change when it comes to the probabilities)
This
together leads to: $$P\left(A\right)=P\left(D\right)=\frac{1}{2}$$
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