What's the probability of choosing a sequence of $4$ numbers, in a particular order? I'm having a dispute with my friend about the following calculation. I was randomly assigned a new phone number that ends in the four digits which represent his birthday ($8479$).  We are trying to calculate what is the probability of those digits being assigned, in that order? 
So far, I've broken it down into two events:


*

*$A=$  Probability of being assigned $8$,$4$,$7$ and $9$ in any order $= \frac1{10000} = 0.0001$

*$B=$  Probability of those numbers being in a specific order $= \frac1{n!} = \frac1{24} \approx 0.041667 $

*$P(A,B)=$  probability that given the $4$ digits are selected, then they
are in the right order $= 0.0001 \times 0.041667 \approx 4.16667 \times 10^{-6}$, or odds of $\frac1{240000}$.


Have I calculated this correctly? If not, what's the right way to calculate it?  I've spend a fair bit of time trying to google this with no luck, and sadly it has been a very  long  time since my university probbability class.
 A: The probability of being assigned $8479$ in that order is $10^{-4}$. 
(We are making the assumption that numbers are assigned "at random" with all strings between $0000$ and $9999$ are equally likely. That may be false. Perhaps "nice" numbers are reserved for businesses willing to pay extra for them.)
A: The calculation of $A$ is wrong. You want to keep the calculation and throw away $A$.
Think of it like this: What is the probability that the first digit is right? What is the probability that the second digit is right? What is the probability that all of them are right?
A: Or to put it another way--there are $10^4$ numbers 'between' 0000 and 9999 and you want ONE.  So it's $\cfrac{1}{10^{4}}$.
A: The probability of four numbers being selected is equal to the number of ways that the four numbers can be arranged in a specific order, divided by the total number of arrangements of those four numbers. Let us first look at the latter: the total number of ways to arrange 8, 4, 7, 9 in any order is $10^4$, since you have place for 4 digits and each place has a choice of 10 different digits (from 0 to 9). Then, if we look at the total number of ways of having the numbers 8, 4, 7, 9 in that order, then we have places for 4 digits, but in each place, there is only 1 digit that we can choose from (i.e. the number that should be assigned to the first place is 8). So, the total arrangements for this is $1^4 =1$.Therefore, the probability of having those four numbers in that order is equal to: $ \frac{1}{10^4} $ = 0.0001. 
