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!

$$\sum\limits_{a_1=0}^6 \sum\limits_{a_2=a_1+1}^7 \sum\limits_{a_3=a_2+1}^8 \sum\limits_{a_4=a_3+1}^9 1 = 210$$
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.
