probability using permutations what is the probability that if we pick n natural numbers then their product's unit digit   contains $2,4,6,8$.and i have no idea how to start ....so please mention all the steps...
i thought that last digit may be $0-9$ and fav cases are $4$..so prob should be $4/10=2/5$..but that is wrong .
correct answer is $(4^n-2^n)/5^n$..so in a way my answer was true for $n=1$. 
please help
 A: We assume that our $n$ numbers are independently chosen, and that each has final digit equally likely to be $0,1,2,3,\dots,9$.
The product ends in $2$, $4$, $6$, or $8$ precisely if (i) the digits $0$ or $5$ are not chosen and (ii) the digits are not all odd.
The probability $0$ or $5$ are never chosen is $\left(\frac{8}{10}\right)^n$. Given that these digits are never chosen, the probability they are all odd is $\left(\frac{1}{2}\right)^n$, so the probability they are not all odd is $1-\left(\frac{1}{2}\right)^n$. Our probability is therefore
$$\left(\frac{8}{10}\right)^n\left(1-\left(\frac{1}{2}\right)^n\right).$$
We can simplify this in various ways. For example, we can multiply through and get $\left(\frac{4}{5}\right)^n-\left(\frac{2}{5}\right)^n$.
Another way: There are $10^n$ equally likely patterns of last digits of our $n$ numbers.  Of these, $8^n$ patterns give us a product whose end digit is neither $0$ nor $5$.  Out of these $8^n$ patterns, $4^n$ give final digit one of $1,3,7,9$. So the number of patterns that give us end digit $2$, $4$, $6$, or $8$ is $8^n-4^n$. 
Our probability is therefore $\frac{8^n-4^n}{10^n}$. 
Remark: It is not possible to have a random variable $X$ which takes on all non-negative integer values with equal probabilities. That is why we interpreted the question to mean that our numbers are chosen to have all final digits equally likely. 
