Question

The numbers $1,2,3,4,.....n$ are arranged in a random order. The probability that the digits $1,2$ and $3$ appear as neighbours in the order named is

• A. $\frac{1}{n}$
• B. $\frac{1}{n-1}$
• C. $\frac{1}{n(n-1)}$
• D. $\frac{1}{n-2}$

Answer 1

The correct option is C $\frac{1}{n(n-1)}$

Total ways of arranging $n$ numbers $=n!$
Create one packet of $1,2,3$
Now total numbers are $n-2$
ways of arranging $n-2$ numbers $=(n-2)!$
$P\left(arranging\right)=(n-2)!/n!=1/(n\ast (n-1\left)\right)$