Probability of drawing cards in ascending order Given 200 cards where each card has a unique number from 1 to 200.
We randomly pick 30 cards (the order we pick them matters). What is the probability the unique numbers of the cards we pick are in ascending order?
 A: Just to elaborate on the comments, I'll give two different arguments.  The first, based purely on symmetry, and the second based on counting.
I. One way to get a uniform distribution is to select an unordered subset (with $30$ elements) and then choose a random permutation of it.  Here, we don't care which unordered subset we choose and we only want one of the $30!$ permutations.  As the permutations are equally probable, the answer is $ {\frac 1{30!}}$.
II. First we count the number of ordered subsets. As noted in the posted solution of @Hamid there are $$\frac {200!}{(200-30)!}$$
How many are in ascending order?  Well as any unordered subset can be put in ascending order in exactly one way there are $$\binom {200}{30}=\frac {200!}{(30!)\times (200-30)!}$$  Hence the probability is the ratio $$\frac {200!}{(30!)\times (200-30)!}\times \frac {(200-30)!}{200!}=\frac 1{30!}$$
A: This answer is for the case when cards are required to be "consecutive", not in "ascending order"
The total number of possibilities is 
$$
\frac{200!}{(200-30)!}
$$
The possible number of ordered choices is $200-30+1 = 171$$^*$. So the probability should be
$$
\frac{200-30+1}{\frac{200!}{(200-30)!}}
$$
$^*$ This comes from $1\to 30$, $2\to 31$, $\cdots$, $171\to200$.
Example: let it $4$ and $2$, your choices are $12,13,14,21,23,24,31,32, 34, 41,42,43$. What you are looking for are $12,23,34$. So the probability is
$$
\frac{4-1+2}{\frac{4!}{(4-2)!}}=\frac{3}{12}
$$
