Probability of having a $4$-digit PIN number with strictly increasing digits What is the probability of having a PIN number (digits $0$-$9$, starting with consecutive zeros allowed) with strictly increasing digits? 
We easily deduce that, if $a_1, a_2, a_3, a_4$ are the respective digits, then $a_1<7, a_2<8$ and $a_3<9$. Moreover, I calculated that there are $7$ choices for $a_1$, ($7-a_1$) choices for $a_2$, ($8-a_2$) choices for $a_3$ and ($9-a_3$) choices for $a_4$ but I don't know how to proceed to the counting with all these variables. Finally, I know that this is ordered sampling without repetition but this doesn't seem to help. Thanks in advance!
 A: You left out an important detail: how is the PIN generated? If you can choose it yourself (as one typically can), then the probability is whatever you want it to be. But let's suppose the PIN has four digits, each of which is chosen independently and uniformly at random.
In that case, the probability of having strictly increasing digits is
$$\frac{|S|}{10^4}$$
where $S = \{0123, 0124, 0125, \ldots, 6789\}$. Now, $S$ corresponds to all possible choices of $4$ different digits from a set of $10$; there are $10 \choose 4$ such choices.
A: For every collection of four distinct digits you choose, precisely one arrangement of those four will be in ascending order.
A: From a set of $\{0, 1, \cdots, 9\}$, choose any subset of 4 numbers. Such subset is in 1-to-1 correspondence with a 4-digit pin with increasing digits.
A: $$\sum\limits_{a_1=0}^6 \sum\limits_{a_2=a_1+1}^7 \sum\limits_{a_3=a_2+1}^8 \sum\limits_{a_4=a_3+1}^9 1 = 210$$
and the total number of possible pins (allowing digit repetition) is of course $10^4$.
Hence:  $P = 0.21$.
Note this is the same value as found by @Théophile and others.
