Problem - 2092
A code consists of four different digits from $1$ to $9$, inclusive. What is the probability of selection a code that consists of four consecutive digits but not necessarily in order? Express your answer as a common fraction.
Answer
$\frac{1}{21}$
Let the smallest number of the $4$ consecutive numbers be $n$. Then $n$ can only be chosen from $1$ to $6$. Furthermore upon fixing the $n$, all these four consecutive numbers are fixed. Hence, the answer to this problem is $$\frac{6\times P_4^4}{P_9^4}=\frac{6\times 4\times 3\times 2\times 1}{9\times 8\times 7 \times 6}=\frac{1}{21}$$