TITLE: Probability of drawing cards in ascending order
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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 $$\frac {200!}{(30!)\times (200-30)!}\times \frac {(200-30)!}{200!}=\frac 1{30!}$$