Question

The numbers $1,2,3,....n$ are arranged in a random order.The probability that the digit $1,2,3,4,5,6,....r(r<n)$ appears as neighbours in that order is

• A. $\frac{1}{n!}$
• B. $\frac{(n-r+1)!}{n!}$
• C. $\frac{r!}{n!}$
• D. None of these

Answer 1

Answer: (B)

$\frac{(n-r+1)!}{n!}$

The exhaustive number of cases $=n!$
Assuming the set of number $1,2,3,...r$ as one the favorable cases $=(n-r+1)!$
$\therefore$ Required probability $=\frac{(n-r+1)!}{n!}$